Question

How much heat is absorbed/released when 35.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) according to the following chemical equation? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) H° = 1168 kJ

Answer 1

Answer:

Amount of energy absorbed = 600.06 kJ

Explanation:

Moles of $NH_3$:-

Mass = 35.00 g

Molar mass of $NH_3$ = 17.031 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{35.00\ g}{17.031\ g/mol}$

$Moles_{NH_3}= 2.055\ mol$

According to the reaction:-

$4 NH_3_{(g)} + 5 O_2_{(g)}\rightarrow 4 NO_{(g)} + 6 H_2O_{(l)}\ H^0 = 1168\ kJ$

Since, enthalpy change is positive, the heat is being absorbed.

4 moles of ammonia on reaction absorbs 1168 kJ of energy

1 mole of ammonia on reaction absorbs 1168/4 kJ of energy

So,

2.055 mole of ammonia on reaction absorbs (1168/4)*2.055 kJ of energy

Amount of energy absorbed = 600.06 kJ