Answer:
[tex]m_{H_2O}=12.9gH_2O[/tex]
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the reaction whereby butane is combusted in the presence of excess oxygen:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
Thus, we can evidence a 2:10 mole ratio of butane to water, and thus, the stoichiometric setup to calculate the mass of produced water is:
[tex]m_{H_2O}=7.49gC_4H_{10}*\frac{1molC_4H_{10}}{52.12gC_4H_{10}} *\frac{10molH_2O}{2molC_4H_{10}}*\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=12.9gH_2O[/tex]
Regards!
