Question

4.The molar enthalpy of combustion of glucose, C6H12O6(s) is –2538.7 kJ/mol.a.Calculate the amount of energy produced by 2.35 mol of glucose.

Answer 1

-5965.9 kJ.

Step-by-step explanation:

What is given?

Molar enthalpy of combustion of glucose (C6H12O6) = -2538.7 kJ/mol.

Moles of glucose = 2.35 moles.

Steb-by-step solution:

Based on the molar enthalpy and the moles, we can do a dimensional analysis to find the amount of energy. Remember that kJ is the unit for energy, so the dimensional analysis will look like this:

`-2538.7\text{ }(kJ)/(mol)\cdot2.35\text{ mol=-5965.9 kJ.}`

The amount of energy produced by 2.35 mol of glucose is -5965.9 kJ.