# C_2H_4 + 3O_2 -> 2CO_2 +2H_2O. If you start with 45 grams of ethylene (C_2H_4), how many grams of carbon dioxide will be produced?

Mar 29, 2017

$\text{No. of mol of ethylene} = \frac{45}{2 \cdot 12 + 4 \cdot 1} \approx 1.607 m o l$
$\text{No. of mol of "CO_2" produced} = 2 \cdot 1.607 = 3.214 m o l$
$\text{Mass of "CO_2} = 3.214 \cdot \left(12 + 2 \cdot 16\right) = 141.416 g$