Answer:
[tex]m_{CO_2}=24.38gCO_2\\m_{H_2O}=8.550gH_2O[/tex]
Explanation:
Hello.
In this case, since the combustion of C7H12 which is probably heptyne (molar mass = 96 g/mol) is:
[tex]C_7H_1_2+10O_2\rightarrow 7CO_2+6H_2O[/tex]
In which the products are carbon dioxide (molar mass = 17 g/mol) and water (18 g/mol) that are in a 7:1 and 6:1 mole ratio with the heptyne, respectively; thus, the yielded mass of both carbon dioxide and water is:
[tex]m_{CO_2}=7.600gC_7H_{12}*\frac{1molC_7H_{12}}{96gC_7H_{12}}*\frac{7molCO_2}{1molC_7H_{12}} *\frac{44gCO_2}{1molCO_2} \\m_{CO_2}=24.38gCO_2\\\\m_{H_2O}=7.600gC_7H_{12}*\frac{1molC_7H_{12}}{96gC_7H_{12}}*\frac{6molH_2O}{1molC_7H_{12}} *\frac{18gH_2O}{1molH_2O} \\m_{H_2O}=8.550gH_2O[/tex]
Best regards.
