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

Step-by-step 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}}=(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=
`(555)/(1)* 0.326=181kJ`

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