Answer:
11.3 g.
Explanation:
Hello there!
In this case, since the combustion of butane is:
[tex]C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O[/tex]
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
[tex]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}[/tex]
Therefore, the resulting mass of water is:
[tex]m_{H_2O}=11.3gH_2O[/tex]
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