Answer:
[tex]m_{CO_2}=45gCO_2[/tex]
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 the molar mass of butane is 58.12 g/mol and that of carbon dioxide is 44.01 g/mol, we obtain the following mass of CO2 product:
[tex]m_{CO_2}=15gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}} *\frac{4molCO_2}{1molC_4H_{10}} *\frac{44.01gCO_2}{1molCO_2} \\\\m_{CO_2}=45gCO_2[/tex]
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