Question

Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen

Answer 1

Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

$C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O$

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

$m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O}$

Therefore, the resulting mass of water is:

$m_{H_2O}=11.3gH_2O$

Best regards!