Question

Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are burned in excess oxygen, how many grams of H2O will be produced?

Answer 1

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of $C_5H_{12}$ = 21.9 g

Molar mass of $C_5H_{12}$ = 72.15 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_5H_{12}$.

$\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles$

Now we have to calculate the moles of $H_2O$.

The balanced chemical reaction will be,

$C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$

From the balanced reaction we conclude that

As, 1 mole of $C_5H_{12}$ react to give 6 moles of $H_2O$

So, 0.3035 moles of $C_5H_{12}$ react to give $0.3035\times 6=1.821$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g$

Therefore, the mass of water produced will be 32.78 grams.