Limiting Reagents

How many grams of NaCl will be produced from 33.0 grams of sodium and 34.0 grams of chlorine? Give your answer to 3 s.f.

Answer explanation

1) Balance the equation:

2Na + Cl₂ --> 2NaCl

2) Calculate the moles of both reactants:

n(Na) = m/M = 33/23 = 1.43

n(Cl₂) = m/M = 34/71 = 0.479

3) Use molar ratio to work out which reagent is in excess and which is limiting:

Na:Cl₂ = 2:1, so 2.86 moles of chlorine required to react with all of the sodium. Chlorine is therefore the limiting reagent.

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

Cl₂:2NaCl has 1:2 ratio, so (0.479x2=) 0.958 moles of NaCl

5) Calculate mass of product from moles and molar mass:

m = nM = 0.958 x 58.5 = 56.043 = 56.0 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

For the reaction below, calculate how many grams of iron oxide will be produced from 10.0 grams of iron and 7.00 grams of oxygen? Give your answer to 3 s.f.

Fe + O₂ --> Fe₂O₃

Answer explanation

1) Balance the equation:

4Fe + 3O₂ --> 2Fe₂O₃

2) Calculate the moles of both reactants:

n(Fe) = m/M = 10/56 = 0.179

n(O₂) = m/M = 7/32 = 0.219

3) Use molar ratio to work out which reagent is in excess and which is limiting:

Fe:O₂ = 4:3, so (0.179/4)*3 = 0.134 moles of oxygen needed to react fully with iron, so iron is the limiting reagent (because we have more oxygen that we need)

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

Fe:Fe₂O₃ in a 4:2 ratio (or 2:1), so (0.179/2)*1 = 0.0895 moles of iron oxide produced.

5) Calculate mass of product from moles and molar mass:

m = nM = 0.0895*160 = 14.32 = 14.3 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Given the reaction C₃H₈ + O₂ --> CO₂ + H₂O, how many grams of carbon dioxide will be produced from 39.0 grams of propane and 11.0 grams of oxygen? Give your answer to 3 s.f.

Answer explanation

1) Balance the equation:

C₃H₈+ 5O₂ --> 3CO₂ + 4H₂O

2) Calculate the moles of both reactants:

n(C₃H₈) = m/M = 39/44 = 0.886

n(O₂) = m/M = 11/32 = 0.344

3) Use molar ratio to work out which reagent is in excess and which is limiting:

C₃H₈:O₂ = 1:5, so (0.886/1)*5 = 4.43 moles of oxygen needed to react fully with propane, so oxygen is the limiting reagent (because we have less oxygen that we need)

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

O₂:CO₂ in a 5:3 ratio, so (0.344/5)*3 = 0.206 moles of carbon dioxide produced.

5) Calculate mass of product from moles and molar mass:

m = nM = 0.206*44 = 9.08 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Sodium chloride can react with silver nitrate in a precipitation reaction. How many grams of silver chloride would be produced from 7.00g of sodium chloride and 95.0g of silver nitrate? Give your answer to 3 s.f.

Answer explanation

1) Balance the equation:

NaCl + AgNO₃ --> AgCl + NaNO₃

2) Calculate the moles of both reactants:

n(NaCl) = m/M = 7/58.5 = 0.120

n(AgNO₃) = m/M = 95/160 = 0.559

3) Use molar ratio to work out which reagent is in excess and which is limiting:

1:1 ratio so we can simply look at the values obtained in step 2 and see that NaCl is our limiting reagent.

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

NaCl & AgCl in a 1:1 ratio so 0.120 moles of AgCl formed.

5) Calculate mass of product from moles and molar mass:

m = nM = 0.120 x (108 + 35.5) = 17.17 = 17.2 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Propane can burn in oxygen to produce carbon dioxide gas. How many grams of carbon dioxide can be produced from 7.00 grams of propane and 98.0 grams of oxygen? Give your answer to 3 significant figures.

Answer explanation

1) Balance the equation:

C₃H₈ + 5O₂ --> 3CO₂ + 4H₂O

2) Calculate the moles of both reactants:

n(C₃H₈) = m/M = 7/44 = 0.159

n(O₂) = m/M = 98/32 = 3.06

3) Use molar ratio to work out which reagent is in excess and which is limiting:

1:5 ratio so 0.159*5 = 0.795 moles of oxygen needed to completely react with propane, so propane is our limiting reagent (because we have 3.06 moles of oxygen and only need 0.795)

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

C₃H₈ is in a 1:3 ratio with CO₂ so 0.159*3 = 0.477 moles of CO₂ produced.

5) Calculate mass of product from moles and molar mass:

m = nM = 0.477 x 44 = 20.988 = 21.0 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Sodium chloride undergoes a precipitation reaction with silver nitrate to produce silver chloride and sodium nitrate. How many grams of silver chloride could be produced from 5.00 g of sodium chloride and 103 g of silver nitrate?

Answer explanation

1) Balance the equation:

NaCl + AgNO₃ --> AgCl + NaNO₃

2) Calculate the moles of both reactants:

n(NaCl) = m/M = 5/58.5 = 0.0855

n(AgNO₃) = m/M = 103/170 = 0.606

3) Use molar ratio to work out which reagent is in excess and which is limiting:

1:1 ratio so we simply look at our values from part 2 and know that the smallest value (NaCl) is our limiting reagent.

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

1:1 ratio again so 0.0855 moles of NaCl produces 0.0855 moles of AgCl

5) Calculate mass of product from moles and molar mass:

m = nM = 0.0855*(108+35.5) = 12.269... = 12.3 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Magnesium hydroxide reacts with hydrochloric acid. How many grams of the resulting salt would be produced from 16.0 grams of the base and 11.0 grams of the acid? Give your answer to 3 s.f.

Answer explanation

1) Balance the equation:

Mg(OH)₂ + 2HCl --> MgCl₂ + 2H₂O

2) Calculate the moles of both reactants:

n(Mg(OH)₂) = m/M = 16/58 = 0.276

n(HCl) = m/M = 11/36.5 = 0.301

3) Use molar ratio to work out which reagent is in excess and which is limiting:

1:2 ratio between base and acid so 0.276 moles of base reacts with 0.552 moles of acid. We only have 0.301 moles of acid, so the HCl is our limiting reagent.

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

HCl:MgCl₂ are in a 2:1 ratio, so 0.301/2 = 0.151 moles of magnesium chloride are produced.

5) Calculate mass of product from moles and molar mass:

m = nM = 0.151 x 95 = 14.2975 = 14.3 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Copper displaces silver from a solution of silver nitrate. How many grams of silver could be obtained by adding 5.00 grams of copper to a solution containing 1.00 grams of silver nitrate? Give your answer to 3 s.f.

Answer explanation

1) Balance the equation:

Cu + 2AgNO₃ --> 2Ag + Cu(NO₃)₂

2) Calculate the moles of both reactants:

n(Cu) = m/M = 5/63.5 = 0.0787

n(AgNO₃) = m/M = 1/170 = 0.00588

3) Use molar ratio to work out which reagent is in excess and which is limiting:

1:2 ratio so 0.0787*2 = 0.157 moles of silver nitrate needed to fully react with the copper. Only have 0.00588 moles so silver nitrate is limiting reagent.

4) Use molar ratio of limiting reagent and product formed to work out moles of product:

Silver nitrate & silver are in 2:2 (1:1) ratio so 0.00588 moles of silver are produced.

5) Calculate mass of product from moles and molar mass:

m = nM = 0.00588 x 108 = 0.635 grams (3 s.f.)

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

2.00 grams of silver are added to 50 cm³ of a 2.35 mol dm^-3 solution of sulfuric acid. What mass of silver sulfate is produced from this reaction?

Answer explanation

Zero grams are produced, as silver is not reactive enough to displace hydrogen from acids and so will not react with sulfuric acid.

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Limiting Reagent

A-Level Chemistry

FormulaeEquations

Hydrogen gas was reacted with excess fluorine to make hydrogen fluoride gas. If 300 cm³ of hydrogen was used, calculate the volume of hydrogen fluoride produced.

Answer explanation

1) Write a balanced equation for the reaction.

H₂ (g) + F₂ (g) --> 2HF (g)

2) Calculate the number of moles of gas

V=nR

0.3 = 24n

n = 0.0125 moles of hydrogen

3) Use molar ratio to work out moles of product produced

H₂ (g) + F₂ (g) --> 2HF (g)

Hydrogen 1:2 Hydrogen fluoride

0.0125 : 0.025 moles HF

4) Calculate volume of product

V= nR

V = 0.025 x 24

V = 0.6 dm³ = 600 cm³

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A-Level Chemistry

FormulaeEquations

Limiting Reagent