Question

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?

Answer 1

the total heat is 33.8 kcal.

Answer 2

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

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of CO = 14 g

Molar mass of CO = 28 g/mol

Putting values in above equation, we get:

`\text{Moles of CO}=(14g)/(28g/mol)=0.5mol`

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

`CO(g)+(1)/(2)O_2(g)\rightarrow CO_2(g)+67.6kCcal`

By Stoichiometry of the reaction:

If 1 mole of CO produces 67.6 kCal of heat

Then, 0.5 moles of CO will produce =
`(67.6kCal)/(1mol)* 0.5mol=33.8kCal` of heat.

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