Calculating quantities of reactants needed, calculating mass of product based on limiting reactant

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

14. Other GCSE chemical calculations

CALCULATING THE QUANTITIES OF REACTANTS NEEDED for a chemical preparation reaction and calculations based on LIMITING REACTANT QUANTITIES

Quantitative chemistry calculations This page explains how to calculate the quantities of reactants needed to prepare a given amount of product. Help in how to theoretically calculate quantities of reactants needed and products formed. Mass or moles of reactants needed for a chemical preparation - fully worked out example calculations and the moles and mass of products that can be theoretically made. How to work out which is the limiting reactant is explained and the resulting theoretical maximum yield calculation in moles and masses of products. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses.  These revision notes and practice questions on how to do chemical calculations and worked examples what amounts of reactants are needed should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

Spotted any careless error? EMAIL query ? comment or request a type of GCSE calculation not covered?

14.2b atom economy calculations  *  14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

14.5 (a) Calculation of quantities required for a chemical reaction

and a brief mention of and

(a) In a chemical preparation reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used but this still requires you to be able to calculate the quantities of chemicals needed.

(b) The reactant chemical that is completely used up is called the limiting reactant because it limits the amount of products after it has all been used up. Therefore you need to be able to explain the effect of a limiting quantity of a reactant on the maximum amount of products it is theoretically possible to obtain in terms of amounts in moles or masses in grams.

BUT, don't forget, you still never get in reality 100% yield.

Calculations of quantities of chemicals required to do a preparation

• Example 1. How much iron and sulfur do you need to heat together to make 20.0 g of iron sulfide

• Atomic masses: Fe = 56 and S = 32

• Balanced equation: Fe + S ==> FeS

• I've set out the solution to the problem in the form of a 'logic' table.

•  This is essentially a reacting mass calculation Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product comments Fe + S ==> FeS You need the balanced chemical equation and in this case its very simple. 56g + 32g ==> 88g basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses 56/88 = 0.6364g + 32/88 = 0.3636 ==> 88g/88 = 1.00g divide by 88 to scale down to 1g of FeS product 20 x 0.6364 = 12.7g + 20 x 0.3636 = 7.3g ==> 20 x 100 = 20.0g then scale by a factor of 20 (for 20g of FeS) Therefore to make 20g of iron sulfide you need 12.7g of iron and 7.3g of sulfur.
• -

• Example 2. How much iron do you need to make 100.0g of iron(III) chloride by passing excess chlorine gas over heated iron filings?

• Atomic masses: Fe = 56 and Cl = 35.5, formula mass of FeCl3 = 162.5 (56 + 3x35.5)

• Balanced equation: 2Fe(s) + 3Cl2(g) ==> 2FeCl3(s)

• Again, I've set out the solution to the problem in the form of a 'logic' table not using moles.

• Be aware that to solve this sort of problem, you only need to pick out the relevant ration.

•  This is essentially a reacting mass calculation Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant and product comments 2Fe ==> 2FeCl3 You need the balanced chemical equation BUT only these bits of the equation are needed to solve the problem, you can ignore the rest of the equation 2 x 56 = 112g ==> 2 x 162.5 = 325g basic reacting mass ratio from the balanced symbol equation and the relevant atomic or formula masses 112/325 = 0.3446g ==> 325/325 = 1.00g divide by 325 to scale down to 1g of FeCl3 product 100 x 0.3446 = 34.46g ==> 100 x 1 = 100.00g then scale by a factor of 100 to make 100g of FeCl3 Therefore to make 100g of iron(III) chloride you need 34.5g of iron (3sf)
• You can do this calculation using the mole concept as follows:

• From the equation 2 moles of iron makes 2 moles of iron(III) chloride.

• Therefore 1 mole Fe makes 1 mol FeCl3

• Atomic mass Fe = 56, Cl = 25.5. Formula mass of FeCl3 = 162.5

• Using the formula on the right twice ...

• mol FeCl3 required = 100/162.5 = 0.615 mol

• Therefore 0.6154 mol of iron is required.

• mass Fe required = 0.6154 x 56 = 34.5 g

• -

• Example 3 A more complex example based on a salt preparation

• Suppose you want to make 50g of the blue hydrated copper(II) sulphate crystals

• Calculation 3(a) based on the mass of copper oxide

• The blue crystals contain water of crystallisation, which must be taken into account in doing the calculation.

• Your reactants are dilute sulphuric acid and the black solid copper(II) oxide.

• You can use copper(II) carbonate, but this is not a pure simple compound and the predictive nature of the calculations will not be as good.

• copper(II) oxide + sulphuric acid ==> copper(II) sulphate + water

• (i) CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l)

• on crystallisation you get the blue hydrated crystals of formula CuSO4.5H2O

• so strictly speaking, after evaporation-crystallisation the overall equation is

• (ii) CuO(s) + H2SO4(aq) + 4H2O(l) ==> CuSO4.5H2O(s)

•    (formula of the blue crystals)
• So, how much copper(II) oxide is needed?

• A 'non-moles' calculation first of all, involving a reacting mass calculation.

• The crucial change overall is CuO ==> CuSO4.5H2O  (mass ratio 80 ==> 250)

• (Reminder: In reacting mass calculations you can often ignore other reactant/product masses, pick out the ratio you need)

• Atomic masses: Cu = 64, S = 32, H = 1, O = 16

• Formula masses are for: CuO = 64 + 16 = 80,   CuSO4 = 64 + 32 + (4 x 16) = 160

• and CuSO4.5H2O = 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250

• The crucial reacting mass ratio is: 80 ==> 250 since formula mole ratio is 1:1 in the equation.

• Therefore, theoretically, to make 50g of the crystals (1/5th of 250),

• you need 1/5th of 80g of copper(II) oxide,

• and 80/5 = 16g, the mass of copper(II) oxide required.

• -

• Calculating the mass of copper oxide needed using moles.

• Formula mass of CuSO4.5H2O  =  250,  formula mass of CuO = 80   (details above)

• 50 g of CuSO4.5H2O = 50/250 = 0.20 mol

• So you need 0.20 mol of CuO  (1 : 1 molar ratio in the balanced symbol equation)

• mass = mol x formula mass   (see triangle on right!)

• mass CuO needed = 0.20 x 80 = 16 g

• The same answer as above, and in my opinion, easier to manage using moles and its better to be able to do mole calculations as well as 'non-mole' reacting mass calculations.

• -

• However, in reality, things are not so simple because the method involves adding excess copper(II) oxide to the dilute sulphuric acid. (see salt preparation method (b) and see the note on 'limiting reactant', the last section of this  page.

• So in practice you would need to use more of the CuO to get anything like 50g of the salt crystals.

• So in calculation 3(b) will look at this preparation from the point of view of how much acid is needed to make the 50 g of copper(II) sulfate crystals.

• -

• Calculation 3(b) based on the volume of dilute sulfuric acid

• There is another way to calculate the quantities required based on the acid and this is the better calculation in which you make the dilute sulfuric acid the limiting reactant and then you can add excess copper oxide (i.e. more than the 16 g of CuO calculated above..

• How much dilute sulphuric acid (of concentration 1 mol dm-3) is required to make 50 g of CuSO4.5H2O?

• Mol = mass in g / formula mass,

• so moles of CuSO4.5H2O required = 50/250 = 0.2 mol (see basics of moles)

• Therefore 0.2 mol of H2SO4 is required (1/5th mol), since the mole ratio CuO : H2SO4 is 1 : 1 reading the equation in terms of moles of reactants and moles of products.

• 1dm3 (1000 cm3) of a 1 mol dm-3 solution of contains 1 mole (by definition - see molarity page)

• Therefore 1/5th of 1 dm3 is required to provide 1/5th of mole of the sulfuric acid.

• so 200 cm3 of 1 mol dm-3 dilute sulphuric acid is required,

• or using just 100 cm3 of 2 mol dm-3 (often labelled 2M on many laboratory reagent bottles)

• You would then add copper(II) oxide in small amounts until no more dissolves in the warm-hot acid and the excess black powder is filtered off. There is no need to weigh out an exact amount of copper oxide.

• If you want just 25g of copper sulfate crystals you would use 100 cm3 of 1 molar sulfuric acid, or 50cm3 of 2 molar sulfuric acid.

• BUT REMEMBER

• (i) in practice, you will NOT get a 100% yield, see calculation below.

• (ii) It would be normal to use excess of the copper oxide, because it is easy to separate by filtration the unreacted oxide to leave a neutral solution of the salt, so you would use more than the 16 g of CuO calculated in part 3(a).

• 3(c) Suppose after carrying out the preparation you finally crystallise 34g of pure the blue crystals of CuSO4.5H2O after weighing the dry product. Losses are inevitable, but now consider the two possible equations, the atom economy and a possible yield in practice.

• (i) CuO(s) + H2SO4(aq) ==> CuSO4(aq) + H2O(l)

• (ii) CuO(s) + H2SO4(aq) +  4H2O(l) ==> CuSO4.5H2O(s)

•    (formula of the blue crystals)
• What is the 'atom economy' of the preparation? (you need to refer to equations (i) and (ii) above.

• Atom economy = useful theoretical products x 100/mass of all reactants

• based on equation (i) Atom economy = 100 x mass CuSO4 / (mass CuO + H2SO4)

• = 100 x 160 / (80 + 98) = 16000/178 = 89.9%

• based on equation (ii) the atom economy is 100% if you include water as a 'reactant', can you see why?

• If you are asked to calculate the atom economy of a reaction you will be given the equation.

• What is the % yield? i.e. comparing what you actually get with the maximum possible, i.e. a 'reality check'!

• % yield = actual mass of product obtained x 100 / theoretical mass from the equation

• % yield = (34 / 50) x 100 =  68% yield

• More examples of % yield and atom economy calculations in section 6.

• and specific sections 14.2a % reaction yield  *  14.2b atom economy

• -

• Example 4

• -

• -

14.5 (b) Calculating limiting quantities of a reactant

Diagram 'borrowed' from one of my 'rates of reaction pages'

The graph above illustrates the effect of a 'limiting reactant'. It might represent a graph when a gas is evolved from e.g. (i) a metal reacting with an acid or (ii) a carbonate dissolving in acid. The little vertical arrow indicates when the reaction has stopped. At this point one of the reactants, the limiting reactant, has all been used up in the reaction. It might be the metal or the carbonate (all the solid dissolved in excess acid) or it might be the acid, whose concentration falls to zero in the presence of excess of the reactant solid.

You need to be able to logically deduce from given quantities of reactants and the chemical equation, which reactant is the limiting reactant?

In a preparation chemical reaction involving two reactants, it is quite common to use an excess of one of the reactants to ensure that all of the other reactant is used up.

The reactant that is completely used up is called the limiting reactant because it limits the maximum amount of products that can be formed i.e. when it is all used up in the reaction, but from the amount of it, you can theoretically calculate the maximum yield of product possible.

The amount of product you can make is directly proportional to the quantity of limiting reactant used i.e. double the mass of the limiting reactant you will double the mass of product. Double the particles present, you double the product particles!

You need to be able to explain the effect of a limiting quantity of a reactant on the amount of products it is possible to make e.g. in terms of amount of product in moles or masses in grams.

How to deduce the limiting reactant and then calculate the maximum quantity of the product

You need to know or be given:

the correctly balanced symbol equation,

The relevant relative atomic masses to work out formula masses

The relationship between moles, mass and atomic/formula mass

mol = mass / (Ar or Mr)    and    mass = mol x (Ar or Mr)

(help triangle on the right, cover over what you want and the rest is how to work it out!)

You can then compare the moles of reactants and see which one (the excess reactant) outweighs the other (the limiting reactant).

AND work out the maximum moles or mass of product from the limiting reactant.

These calculations do assume 100% yield, but so in reality (% reaction yield and theoretical yield calculations)

Example 1a. Deducing the limiting reactant and calculating the maximum quantity of the product

Making iron sulfide

If 40 g of sulfur is heated with 60g of iron filings to make iron sulfide, which is the limiting reactant? and what is the maximum mass of iron sulfide that can be made?

The reaction is: Fe + S ===>  FeS;  atomic masses: Fe = 56;  S = 32

(i) Calculating the limiting reactant

moles = mass in g / formula mass

moles S = 40 / 32 = 1.25 mol S

moles Fe = 60 / 56 = 1.07 mol Fe

Since one mole of iron reacts with one mole of sulfur, there is insufficient iron to react with all of the sulfur used in the experiment, therefore the iron is the limiting reactant.

(ii) Calculating the maximum yield from the limiting reactant.

This part of the calculation MUST be based on the limiting reactant (1.07 mol Fe)

Formula mass iron sulfide FeS = 56 + 32 = 88

From the reaction equation, reading in molar quantities, one mole of iron forms one mole of iron sulfide.

mass FeS = moles of FeS  x  formula mass FeS

mass FeS = 1.07 x 88 = 94.2 g

Example 1b. Deducing the limiting reactant and calculating the maximum quantity of the product

Making oxides by heating a metal in air

In this case the metal is automatically the limiting reactant if it is fully exposed to excess of air, effectively excess oxygen.

In these examples I have omitted the experiment details and details of weighings.

Atomic masses for these three examples: Al = 27,  Mg = 24,  Fe = 56, O = 16

Example (i) Heating magnesium in air in a crucible

2Mg  +  O2  ===>  2MgO

If you burn 3.0 g of magnesium what is the maximum mass of magnesium oxide that can be formed?

mol Mg = 3/24 = 0.125 mol

In the equation 1 mol of Mg gives 1 mol of MgO (formula mass = 24 + 16 = 40)

Therefore maximum mass of MgO product = 0.125 x 40 = 5.0 g MgO

You can also deduce that 2.0 g of oxygen was consumed in the reaction.

Example (ii) Heating aluminium air in a crucible

2Al  +  3O2  ===>  2Al2O3

What mass of aluminium do you need to make 20.0 g of aluminium oxide?

Formula mass of Al2O3 = (2 x 27) + (3 x 16) = 102

mol Al2O3 required = 20.0/102 = 0.196 mol

In the equation 2 mol of Al gives 2 mol Al2O3

This is a 1 : 1 ration so mol Al = mol Al2O3

Therefore mass of Al needed = 0.196 x 27 = 5.29 g (3 sf, 2 dp)

Example (iii) Heating iron in a stream of steam

When excess steam is passed over iron at 450oC the oxide formed has the formula Fe3O4.

This means the iron is the limiting reactant.

The equation is: 3Fe  +  2O2  ===>  Fe3O4

If 14.0g of iron is heated in excess steam what mass of the iron oxide is formed?

Atomic mass Fe = 56, formula mass of Fe3O4 = {(3 x 56) + (4 x 16)} = 168 + 64 = 232

mol Fe = 14.0 / 56 = 0.25 mol

Now according to the equation above 3 mol of Fe gives 1 mol of Fe3O4

Therefore mol Fe3O4 = mol Fe / 3 = 0.25 / 3 = 0.08333

Therefore mass of Fe3O4 = 0.08333 x 232 = 19.3 g (3 sf, 1 dp)

You can work the calculation the other way and work out the amount of iron needed to make a given mass of Fe3O4.

e.g. How much iron is needed to make 50.0 g of Fe3O4?

From the equation you need 3 mol of iron to make 1 mol of Fe3O4

mol of Fe3O4 = 50.0 / 232 = 0.2155

mol Fe needed = 3 x mol Fe3O4

mol Fe needed = 3 x 0.2155 = 0.6465

mass of Fe required = 0.6465 x 56 = 36.2 g

Example 2. Deducing the limiting reactant and calculating the maximum quantity of the product

Making copper sulfate from copper oxide and sulfuric acid

If 10 g of copper oxide (CuO) is added to 50 cm3 of sulfuric acid of concentration 2.0 mol/dm3, deduce:

(a) which is the limiting reactant?

(a) What is the maximum amount of copper sulfate (as CuSO4) you could theoretically make?

The reaction is: CuO(s)  +  H2SO4(aq)  ===> CuSO4(aq)  + H2O(l)

Atomic masses: Cu = 64; O = 16; H = 1; S = 32 (but you might not need all of these!)

(a) Deducing the limiting reactant

Formula mass copper sulfate = 64 + 16 = 80

moles = mass in g / formula mass

mol CuO = 10 / 80 = 0.125 mol

mol H2SO4 = molarity x volume (in dm3)    (dm3 = cm3 / 1000)

mol H2SO4 = 2.0 x 50 / 1000 = 0.10 mol

From the equation, reading in molar quantities, one mole of copper oxide reacts with one mole of sulfuric acid to make one mole of copper sulfate.

Clearly the copper oxide is in excess, so the sulfuric acid is the limiting reactant, because you would need 0.125 moles of the acid (only got 0.10 mol) to react with all 10 g (0.125 mol) of the copper oxide.

In the actual preparation of the copper sulfate salt crystals you would filter off the excess copper oxide, so this is a viable recipe for making copper(II) sulfate crystals..

(b) Calculating the theoretical maximum yield of copper sulfate from the limiting reactant

This part of the calculation MUST be based on the limiting reactant (0.10 mol H2SO4)

From the equation, reading in molar quantities:

one mole of sulfuric acid makes one mole of copper sulfate (0.1 mol).

Formula mass copper sulfate = 64 + 32 + (4 x 16) = 160

mass = moles x formula mass = 0.10 x 160 = 16g of CuSO4

Strictly speaking the actually product is the 'blue' hydrated copper sulfate crystals, formula CuSO4.5H2O

This has a formula mass of 160 + (5 x 18) = 250

Therefore the maximum possible mass of 'blue' crystals is 0.1 x 250 = 25g of CuSO4.5H2O

Example 3. Deducing the limiting reactant and calculating the maximum quantity of the product

Making hydrogen from magnesium and hydrochloric acid

Suppose we take a strip of magnesium weighing 0.36g and dissolve it in 50 cm3 of hydrochloric acid of molarity 1.0 mol/dm3..

The equation is: Mg(s)  +  2HCl(aq)  ===>  MgCl2(aq)  +  H2(g)   and  Ar(Mg) = 24

(a) Which is the limiting reactant?

To calculate the limiting reactant we need to compare the number of moles of each reactant, AND then the relative reaction mole ratio to see which one is in excess - take care, you must not automatically just compare the actual number of moles of each reactant, you must consult the equation too!

mol Mg = mass Mg / Ar(Mg) = 0.36 / 24 = 0.015 mol Mg

moles = molarity x volume (in dm3)

mol HCl = [molarity HCl] x [volume of HCl(aq) in dm3]   (dm3 = cm3 / 1000)

mol HCl = 1.0 x (50 / 1000) = 0.05 mol HCl

According to the equation each mole of Mg requires two moles of HCl to complete the reaction.

Therefore 0.015 mol Mg requires 0.030 mol of HCl to react completely.

Since the moles of HCl required (0.03) is exceeded by the moles of HCl used (0.05),

the magnesium must be the limiting reactant

(b) What is the maximum volume of hydrogen gas you can make in this experiment from the limiting reactant?

This part of the calculation MUST be based on the limiting reactant (0.015 mol Mg)

Volume of 1 mol of gas = 24 dm3 at RTP (room temperature and pressure)

From the equation, reading in molar quantities, each mole of magnesium gives one mole of hydrogen gas,

so mol Mg used = mol H2 formed = 0.015 (limiting moles)

volume H2 = mol H2 x 24

= 0.015 x 24 = 0.36 dm3 (360 cm3)

Example 4. Deducing the limiting reactant and calculating the maximum quantity of the product

Preparing carbon dioxide from limestone and acid

2g of calcium carbonate granules (limestone) was dissolved in 20 cm3 of hydrochloric acid of concentration 0.50 mol/dm3.

The reaction is: CaCO3(s)  +  2HCl(aq)  ===>  CaCl2(aq)  + H2O(l)  +  CO2(g)

Atomic masses: Ca = 40; C = 12; O = 16; H = 1, Cl = 35.5; but you might not need all these Ar values!

(a) Which is the limiting reactant?

Formula mass CaCO3 = 40 + 12 + (3 x 16) = 100

mol CaCO3 = 2 / 100 = 0.02

mol HCl = [molarity HCl] x [volume of HCl(aq) in dm3]   (dm3 = cm3 / 1000)

mol HCl = 0.5 x 20 / 1000 = 0.01 mol

From the equation one mole of calcium carbonate requires two moles of hydrochloric acid to complete the reaction. Therefore the 0.02 mol of CaCO3 needs 0.04 mol of HCl to react completely.

Since only 0.01 mol of HCl was used in the mixture, clearly the calcium carbonate is in great excess and the hydrochloric acid is the limiting reactant.

Note: At the end of the reaction, indicated by no more 'fizzing', you would see white unreacted granules of limestone at the bottom of the flask.

(b) What is the maximum volume of carbon dioxide gas that can be obtained using this mixture from the limiting reactant?

This part of the calculation MUST be based on the limiting reactant (0.01 mol HCl)

From the equation, reading in molar quantities, two moles of hydrochloric acid produces one mole of carbon dioxide (so the moles of HCl will only make half the amount of moles of CO2).

Therefore mole CO2 = mol HCl / 2 = 0.01 / 2 = 0.005

volume CO2 = mol CO2 x 24 dm3

= 0.005 x 24 = 0.12 dm3  (or 0.12 x 100 = 120 cm3)

14.2b atom economy calculations  *  14.3 dilution of solutions calculations  *  14.4 water of crystallisation calculations

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

Above is typical periodic table used in GCSE science-chemistry specifications in doing chemical calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

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