Question

How many grams of H2O will be produced from the complete combustion of 26.7 g of butane (C4H10)?

Answer 1

$2C_4H_{10}+ 13H_2O \rightarrow 8CO_2 + 10H_2O$

From the chemical reaction above, 2 moles of Butane produces 10 moles of water.

Therefore, we can write;

$\begin{aligned} 2\textsf{ moles of }C_4H_{10} &= 10 \textsf{ moles of }H_2O\\ 2(54\ g) &= 10(18\ g)\\ \\ 108\textsf{ g of }\ C_4H_{10} &= 180\textsf{ g of } H_2O\\ 26.7\textsf{ g of }\ C_4H_{10} &= x\textsf{ g of } H_2O\\ \\ \textsf{mass of }H_2O (x\ g) &= \dfrac{26.7×180}{108}\\ \\&=44.5\ g \end{aligned}$

$\therefore$ 44.5g grams of $H_2O$ will be produced from the complete combustion of 26.7g of butane.