Stoichiometry Chemistry Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

medium
How many grams of O2\text{O}_2 are needed to completely burn 16.016.0 g of CH4\text{CH}_4? (CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O})

Solution

  1. 1
    Moles of CH4=16.016.04=0.998 mol≈1.00 mol\text{CH}_4 = \frac{16.0}{16.04} = 0.998\,\text{mol} \approx 1.00\,\text{mol}.
  2. 2
    From the equation: 1 mol CH4\text{CH}_4 requires 2 mol O2\text{O}_2. So moles of O2=2×1.00=2.00 mol\text{O}_2 = 2 \times 1.00 = 2.00\,\text{mol}.
  3. 3
    Mass of O2=2.00×32.00=64.0 g\text{O}_2 = 2.00 \times 32.00 = 64.0\,\text{g}.

Answer

64.0 g of O264.0\,\text{g of O}_2
Stoichiometry uses mole ratios from balanced equations to convert between amounts of reactants and products. The steps are: grams → moles → mole ratio → moles → grams.

About Stoichiometry

The branch of chemistry that uses balanced chemical equations and mole ratios to calculate the precise quantities of reactants consumed and products formed in chemical.

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More Stoichiometry Examples

Example 2 hard

How many grams of [formula] can be produced from [formula] g of [formula] and excess [formula]? ([fo

Example 3 medium

How many moles of [formula] are produced when [formula] mol of [formula] reacts with excess [formula

Example 4 medium

How many grams of [formula] form from [formula] mol of [formula] reacting with excess sodium? ([form

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