Mass–mass stoichiometry: Class 11 practice

What you should know first

• Mole concept
• Balancing equations

The idea behind mass–mass stoichiometry

A balanced equation gives the mole ratio of reactants and products. Convert given mass to moles, use the ratio, then convert to the mass asked for.

Step-by-step method

1. Balance the equation.
2. Moles of given = mass ÷ molar mass.
3. Use the mole ratio.
4. Mass of product = moles × molar mass.

Worked example 1 (Basic)

Zn + 2HCl → ZnCl₂ + H₂. What mass of H₂ forms from 292.5 g of Zn?

1. n(Zn) = 292.5/65 = 4.5 mol
2. n(H₂) = 4.5 × 1/1 = 4.5 mol
3. m(H₂) = 4.5 × 2 = 9 g
4. Answer: 9 g

Worked example 2 (Standard)

N₂ + 3H₂ → 2NH₃. What mass of NH₃ forms from 14 g of N₂?

1. n(N₂) = 14/28 = 0.5 mol
2. n(NH₃) = 0.5 × 2/1 = 1 mol
3. m(NH₃) = 1 × 17 = 17 g
4. Answer: 17 g

Worked example 3 (Challenge)

2KClO₃ → 2KCl + 3O₂. What mass of O₂ forms from 612.5 g of KClO₃?

1. n(KClO₃) = 612.5/122.5 = 5 mol
2. n(O₂) = 5 × 3/2 = 7.5 mol
3. m(O₂) = 7.5 × 32 = 240 g
4. Answer: 240 g

Common mistakes and misconceptions

• Using mass ratios directly
• Unbalanced equation

Practice questions

1. (Basic) 2KClO₃ → 2KCl + 3O₂. What mass of O₂ forms from 183.75 g of KClO₃?
2. (Basic) N₂ + 3H₂ → 2NH₃. What mass of NH₃ forms from 84 g of N₂?
3. (Standard) Zn + 2HCl → ZnCl₂ + H₂. What mass of H₂ forms from 325 g of Zn?
4. (Standard) Zn + 2HCl → ZnCl₂ + H₂. What mass of H₂ forms from 65 g of Zn?
5. (Challenge) Zn + 2HCl → ZnCl₂ + H₂. What mass of H₂ forms from 227.5 g of Zn?
6. (Challenge) C + O₂ → CO₂. What mass of CO₂ forms from 48 g of C?

Answer key with full working

• Q1: n(KClO₃) = 183.75/122.5 = 1.5 mol → n(O₂) = 1.5 × 3/2 = 2.25 mol → m(O₂) = 2.25 × 32 = 72 g → Answer: 72 g
• Q2: n(N₂) = 84/28 = 3 mol → n(NH₃) = 3 × 2/1 = 6 mol → m(NH₃) = 6 × 17 = 102 g → Answer: 102 g
• Q3: n(Zn) = 325/65 = 5 mol → n(H₂) = 5 × 1/1 = 5 mol → m(H₂) = 5 × 2 = 10 g → Answer: 10 g
• Q4: n(Zn) = 65/65 = 1 mol → n(H₂) = 1 × 1/1 = 1 mol → m(H₂) = 1 × 2 = 2 g → Answer: 2 g
• Q5: n(Zn) = 227.5/65 = 3.5 mol → n(H₂) = 3.5 × 1/1 = 3.5 mol → m(H₂) = 3.5 × 2 = 7 g → Answer: 7 g
• Q6: n(C) = 48/12 = 4 mol → n(CO₂) = 4 × 1/1 = 4 mol → m(CO₂) = 4 × 44 = 176 g → Answer: 176 g

Syllabus note

This skill is listed under "Some Basic Concepts of Chemistry" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

• Some Basic Concepts of Chemistry chapter notes (Class 11)
• Molarity of solutions
• Limiting reagent
• Empirical and molecular formula
• Theoretical and percentage yield
• Formula from combustion data
• Quantum numbers and orbitals
• Hydrogen spectrum and Rydberg formula
• Hybridisation and molecular shape (VSEPR)
• Hess's law and enthalpy of reaction
• All Class 11 practice

Frequently asked questions

Which chapter is mass–mass stoichiometry in?

"Some Basic Concepts of Chemistry", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.