Question

10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and oxygen then react to yield NO gas and water vapor, what is the final density of the gas mixture inside the vessel at 23°C?

Answer 1

Answer:

The density is 3g/L

Explanation:

The reaction that occurs in the vessel is:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

10,0g of NH₃ are:

$10,0g * \frac{1mol}{17,031g} = 0,587 moles$

6,50 g of O₂ are:

$6,50g * \frac{1mol}{32g} = 0,203 moles$

For a complete reaction of O₂ there are necessaries:

$0,203 mol * \frac{4molNH_{3}}{5molO_{2}}= 0,163 moles of NH_{3}$

O₂ is limiting reactant. The excess moles of NH₃ are:

0,587 - 0,163 = 0,424 moles of NH₃

These moles are:

$0,424mol * \frac{17,031g}{1mol} =$ 7,22g of NH₃

Knowing O₂ is limiting reactant, mass of NO and H₂O are:

$0,203molO_{2}*\frac{4molNO}{5molO_{2}}*\frac{30,01g}{1molNO} =$ 4,87g of NO

$0,203molO_{2}*\frac{6molH_{2}O}{5molO_{2}}*\frac{18,02g}{1molH_{2}O} =$ 4,39g of H₂O

The total mass is: 7,22g + 4,87g + 4,39g = 16,48g ≡ 16,5g

-The same mass add in the first. By matter conservation law-

As vessel volume is 5,50L, density is:

16,5g/5,50L = 3g/L

I hope it helps!