Question

g determine the mass of products formed by the combustion of 7.600 g of a compound with empirical formula c7h12

Answer 1

`m_(CO_2)=24.38gCO_2\\m_(H_2O)=8.550gH_2O`

Step-by-step explanation:

Hello.

In this case, since the combustion of C7H12 which is probably heptyne (molar mass = 96 g/mol) is:

`C_7H_1_2+10O_2\rightarrow 7CO_2+6H_2O`

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:

`m_(CO_2)=7.600gC_7H_(12)*(1molC_7H_(12))/(96gC_7H_(12))*(7molCO_2)/(1molC_7H_(12)) *(44gCO_2)/(1molCO_2) \\m_(CO_2)=24.38gCO_2\\\\m_(H_2O)=7.600gC_7H_(12)*(1molC_7H_(12))/(96gC_7H_(12))*(6molH_2O)/(1molC_7H_(12)) *(18gH_2O)/(1molH_2O) \\m_(H_2O)=8.550gH_2O`

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