Question

Combustion of glucose takes place according to the equation
${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{+}\text{6}{\text{O}}_{\text{2}}\to \text{6}{\text{CO}}_{\text{2}}\text{+}\text{6}{\text{H}}_{\text{2}}\text{O;}\text{ΔH}\text{=}-\text{72}\text{kcal/mol}\text{.}$
How much energy will be released by the combustion of $\text{1}\text{.6}\text{g}$ of glucose (Molecular mass of glucose = $\text{180}\text{g/mol}$ )?

• A. 0.064 kcal
• B. 0.64 kcal
• C. 6.4 kcal
• D. 64 kcal

Answer 1

Answer: (B)

Combustion of glucose takes place according to the equation
${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{+}{\text{6O}}_{\text{2}}\to {\text{6CO}}_{\text{2}}\text{+}{\text{6H}}_{\text{2}}\text{O;}\text{ΔH}\text{=}\text{-72}\text{kcal}$
$\Delta H$ per $1.6g=\frac{72\times 1.6}{180}=0.64kcal$