Question

How much heat is absorbed during production of 112g of NO by the combination of nitrogen and oxygen? . N2(g)+O2(g)→2NO(g), ΔH = + 43 kcal/mol

Answer 1

It is calculated as follows:

112g of NO ( 1 mo NO/Molar mass of NO ) x 43Kcal/ 1 mol NO

For these type of problems, you want to convert the amount given to moles first since kcal is over moles. 

Answer 2

Answer : The heat absorbed during the reaction is 160.4 kcal

Explanation :

First we have to calculate the number of moles of NO.

$\text{Moles of }NO=\frac{\text{Mass of }NO}{\text{Molar mass of }NO}$

Molar mass of NO = 30 g/mole

$\text{Moles of }NO=\frac{112g}{30g/mole}=3.73mole$

Now we have to calculate the heat absorbed during the reaction.

$\Delta H=\frac{q}{n}$

or,

$q=\Delta H\times n$

where,

$\Delta H$ = enthalpy change = +43 kcal/mol

q = heat absorb = ?

n = number of moles of NO = 3.73 mol

Now put all the given values in the above formula, we get:

$q=(+43 kcal/mol)\times (3.73mol)=+160.4kcal$

Therefore, the heat absorbed during the reaction is 160.4 kcal