Answer:
0.99 g
Explanation:
Using ideal gas equation to calculate the moles of water as:-
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 712 mmHg
V = Volume of the gas = 5.00 L
T = Temperature of the gas = [tex]380^oC=[380+273]K=653\ K[/tex]
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]712mmHg\times 5.00L=n\times 62.3637\text{L.mmHg}mol^{-1}K^{-1}\times 653K\\\\n=\frac{712\times 5.00}{62.3637\times 653}=0.0874mol[/tex]
Moles of water = 0.0874 moles
From the reaction,
[tex]4NO_2_{(g)}+6H_2O_{(g)}\rightarrow 7O_2_{(g)}+4NH_3_{(g)}[/tex]
6 moles of water forms 4 moles of ammonia
1 mole of water forms 4/6 moles of ammonia
0.0874 moles of water forms (4/6)*0.0874 moles of ammonia
Moles of ammonia formed = 0.0583 moles
Molar mass of ammonia = 17.031 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]0.0583= \frac{Mass}{17.031\ g/mol}[/tex]
[tex]Mass\ of\ ammonia= 0.99 g[/tex]
