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:
[tex]10,0g * \frac{1mol}{17,031g} = 0,587 moles[/tex]
6,50 g of O₂ are:
[tex]6,50g * \frac{1mol}{32g} = 0,203 moles[/tex]
For a complete reaction of O₂ there are necessaries:
[tex]0,203 mol * \frac{4molNH_{3}}{5molO_{2}}= 0,163 moles of NH_{3}[/tex]
O₂ is limiting reactant. The excess moles of NH₃ are:
0,587 - 0,163 = 0,424 moles of NH₃
These moles are:
[tex]0,424mol * \frac{17,031g}{1mol} =[/tex] 7,22g of NH₃
Knowing O₂ is limiting reactant, mass of NO and H₂O are:
[tex]0,203molO_{2}*\frac{4molNO}{5molO_{2}}*\frac{30,01g}{1molNO} =[/tex] 4,87g of NO
[tex]0,203molO_{2}*\frac{6molH_{2}O}{5molO_{2}}*\frac{18,02g}{1molH_{2}O} =[/tex] 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!
