Question

Qmnole Ha5. How much Heat will be released when 10 grams of Ammonia (NH3) reacts withexcess O2 in the following equation? 4NH3 + 502-> 4NO + 6H2OH = 1, 200 kJ =

Answer 1

176.16 kJ

Step-by-step explanation

Given:

Mass of NH3 = 10 grams

Equation: 4NH3 + 5O2-> 4NO + 6H2O ΔH = 1, 200 kJ

What to find:

The heat released when 10 grams of NH3 reacts with excess O2.

Step-by-step solution:

The first step is to convert the 10 grams of NH3 that reacts to mole using the mole formula

`Mole=\frac{Mass}{Molar\text{ }mass}`

Molar mass of NH3 = 17.031 g/mol

`Mole=\frac{10g}{17.031\text{ }g\text{/}mol}=0.5872\text{ }mol`

Now, you can use the mole of NH3 and the given equation to calculate the heat released.

From the equation;

4 moles of NH3 released 1,200 kJ of heat

Therefore, 0.5872 moles of NH3 will release

`\frac{0.5872\text{ }mol\text{ }NH_3}{4\text{ }mol\text{ }NH_3}*1200\text{ }kJ=176.16\text{ }kJ`

Therefore, the heat released when 10 grams of NH3 reacts with excess O2 is 176.16 kJ