Question

if we react 30 grams of ethane ($c_2h_6$) with 320 grams of oxygen, how many grams of carbon dioxide would we make?
$2c_2h_6(g) + 7o_2(g) \
ightarrow 4co_2(g) + 6h_2o(g)$
options: 1 grams, 88 grams, 2 grams, 44 grams

Explanation:

Step1: Calculate molar masses

Molar mass of $\text{C}_2\text{H}_6$: $2\times12 + 6\times1 = 30\ \text{g/mol}$
Molar mass of $\text{O}_2$: $2\times16 = 32\ \text{g/mol}$
Molar mass of $\text{CO}_2$: $12 + 2\times16 = 44\ \text{g/mol}$

Step2: Find moles of reactants

Moles of $\text{C}_2\text{H}_6$: $\frac{30\ \text{g}}{30\ \text{g/mol}} = 1\ \text{mol}$
Moles of $\text{O}_2$: $\frac{320\ \text{g}}{32\ \text{g/mol}} = 10\ \text{mol}$

Step3: Identify limiting reactant

From reaction: $2\ \text{mol}\ \text{C}_2\text{H}_6$ needs $7\ \text{mol}\ \text{O}_2$
For $1\ \text{mol}\ \text{C}_2\text{H}_6$, required $\text{O}_2$: $\frac{7}{2}\times1 = 3.5\ \text{mol}$
Available $\text{O}_2$ (10 mol) > required, so $\text{C}_2\text{H}_6$ is limiting.

Step4: Relate moles to product

$2\ \text{mol}\ \text{C}_2\text{H}_6$ produces $4\ \text{mol}\ \text{CO}_2$
$1\ \text{mol}\ \text{C}_2\text{H}_6$ produces $\frac{4}{2}\times1 = 2\ \text{mol}\ \text{CO}_2$

Step5: Calculate mass of $\text{CO}_2$

Mass = moles $\times$ molar mass
Mass = $2\ \text{mol} \times 44\ \text{g/mol} = 88\ \text{g}$

Answer:

88 grams