The combustion of 14 grams of CO, according to the reaction CO(g) + ½O2(g) → CO2(g) + 67.6 kcal, gives off how much heat?

Respuesta :

the total heat is 33.8 kcal.

Answer: The amount of heat released by the combustion of 14 g of CO is 33.8 kCal

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of CO = 14 g

Molar mass of CO = 28 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of CO}=\frac{14g}{28g/mol}=0.5mol[/tex]

The chemical reaction for the combustion of CO follows the equation:

[tex]CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)+67.6kCcal[/tex]

By Stoichiometry of the reaction:

If 1 mole of CO produces 67.6 kCal of heat

Then, 0.5 moles of CO will produce = [tex]\frac{67.6kCal}{1mol}\times 0.5mol=33.8kCal[/tex] of heat.

Hence, the amount of heat released by the combustion of 14 g of CO is 33.8 kCal