Answer:
[tex]m_{H_2O}=2.22gH_2O[/tex]
Explanation:
Hello!
In this case, since the combustion of butane is:
[tex]C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O[/tex]
As there is an excess of oxygen, we can compute the mass of water by simply using the molar masses of butane and water and the 1:5 mole ratio between them as shown below:
[tex]m_{H_2O}=1.43gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}} *\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O} \\\\m_{H_2O}=2.22gH_2O[/tex]
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