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Stoichiometry

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Stoichiometry

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Calculations

Created by

Alice Low
Alice Low
Singapore

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Stoichiometry
 

StoichiometryOnline version

Calculations

by Alice Low
1

What is the molarity of a solution prepared by dissolving 5.0 grams of NaCl in enough water to make 250 mL of solution? (Molar mass of NaCl = 58.5 g/mol)

2

A sample of ore is 80% pure. If you have 250 grams of the ore, what is the mass of the pure substance?

Choose one or more answers

3

If the theoretical yield of a reaction is 120 grams and the actual yield is 78 grams, what is the % yield?

4

If the actual yield of a product is 60 grams and the percentage yield is 75%, what is the theoretical yield?

5

What is the mass of KBr required to make 500mL of 0.168 M solution? (Molar mass of KBr = 119 g/mol)

6

If 10 grams of H₂ are reacted with 52 grams of O₂, how much water is produced? 2H₂ (g) + O₂ (g) → 2H₂O (g)

7

How many moles and mass of HCl are needed to react completely with 45 g of Zn? Zn + 2HCl → ZnCl₂ + H₂ (Zn = 65.4, H = 1 and Cl = 35.5 )

8

Which of the following is the excess reagent when 1,5 moles of Al react with 6 moles of HCl? Unbalanced: Al (s) + HCl (aq) → AlCl₃ (aq) + H₂ (g)

9

What is the theoretical yield of H₂ in dm³ at rtp when 4 moles of K react with 3 moles of H₂SO₄? Unbalanced: K (s) + H₂SO₄ (aq) → K₂SO₄ (aq) + H₂ (g)

10

What is the concentration of NaOH in a solution if 40.0 mL of 0.500 M H₂SO₄ is required to neutralize 100.0 mL of NaOH solution? Unbalanced: NaOH (aq) + H₂SO₄ (aq) → Na₂SO₄ (aq) + H₂O (l)

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Molarity = moles of solute / liters of solution = (5.0 g / 58.5 g/mol) / 0.25 L = 0.342 M

Pure substance = % purity × total mass = 0.80 × 250 g = 200 grams

% Yield = (actual yield / theoretical yield) × 100 = (78 g / 120 g) × 100 = 65%

Theoretical yield = actual yield / % yield × 100 = 60 g / 75% × 100 = 80 grams

Mass = (Concentration x liters of solution) x molar mass = (0.168 M x 0.5 L) x 119 g/mol = 10 g

Moles of H₂ = 10/2 = 5 mol. Moles of O₂ = 52/32 = 1.625 mol. H₂ is in excess (limiting reagent is O₂). Moles of H₂O = 1.625 × 2 = 3.25 mol. Mass of H₂O = 3.25 × 18 = 58.5 g.

Moles of Zn = 45/65.4 = 0.688 mol. Mass of HCl = 0.688 x 2 x36.5 = 50.23 g Mole ratio of Zn: HCl = 1:2

2Al (s) + 6HCl (aq) → 2AlCl₃ (aq) + 3H₂ (g). HCl is in excess (1.5 mole left unreacted).

2K (s) + H₂SO₄ (aq) → K₂SO₄ (aq) + H₂ (g). Limiting reagent is K. By mole ratio K: H₂SO₄: H₂ = 2: 1: 1 = 4: 2: 2. Theoretical yield of H₂ at rtp = 2 x 24 = 48 dm³.

2NaOH (aq) + H₂SO₄ (aq) → Na₂SO₄ (aq) + 2H₂O (l). Moles = 0.5 x 0.04 = 0.02 mol. Mole ratio of NaOH:H₂SO₄ = 2:1 Molarity of NaOH = (0.02 x 2)/0.1 = 0.400 M

 
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