Question

The complete combustion of ethanol, C₂H₅OH (FW = 46.0 g/mol), proceeds as follows: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$; ΔH = −555 kJ What is the enthalpy change for combustion of 15.0 g of ethanol?

Answer 1

Answer: 181 kJ

Explanation:

The balanced chemical reaction is;

$C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2Ol)$ $\Delta H=-555jJ$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{15.0g}{46.0g/mol}=0.326moles$

According to stoichiometry:

1 mole of $C_2H_5OH$ on complete combustion give= 555 kJ

Thus 0.326 moles of $C_2H_5OH$ on complete combustion give=$\frac{555}{1}\times 0.326=181kJ$

Thus the enthalpy change for combustion of 15.0 g of ethanol is 181 kJ