The reaction below will occur in a gaseous system at STP: 6 NO (g) + 4 NH3 (g) —> 5 N2 (g) + 6 H20 (g)
What volume of nitrogen gas will be produced from 15.0 L of nitrogen monoxide gas?
12.5 L
18.0L
15.0 L
22.5 L

V = 12.5 L

Explanation:

Given data:

Volume of NO = 15.0 L

Temperature and pressure = standard

Volume of nitrogen gas produced = ?

Solution:

Chemical equation:

6NO + 4NH₃    →     5N₂ + 6 H₂O

Number of moles of NO:

PV = nRT

n = PV/RT

n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K

n = 15.0 atm.L / 22.43 atm.L /mol

n = 0.67 mol

now we will compare the moles of No and nitrogen gas.

              NO           :         N₂

               6             :          5

              0.67         :         5/6×0.67 = 0.56

Volume of nitrogen gas:

 PV = nRT

1 atm × V = 0.56 mol ×  0.0821 atm.L /mol.K × 273.15 K

V = 12.5 atm.L / 1 atm

V = 12.5 L

these nuts