The answer is: volume of ammonia gas is 7.4 L.
Chemical reaction: 6NO + 4NH₃ → 5N₂ + 6H₂O.
m(NO) = 15 g; mass of nitrogen(II) oxide.
M(NO) = 30 g/mol; molar mass of nitrogen(II) oxide.
V(NH₃) = ?
n(NO) = 15 g ÷ 30 g/mol.
n(NO) = 0.5 mol; amount of nitrogen(II) oxide.
From chemical reaction: n(NO) : n(NH₃) = 6 : 4.
0.5 mol : n(NH₃) = 6 : 4.
n(NH₃) = 0.33 mol; amount of ammonia.
Vm = 22.4 L/mol; molar volume at STP.
V(NH₃) = 0.33 mol · 22.4 L/mol..
V(NH₃) = 7.4 L.
Answer: 7.47 litres
Explanation:
Mass of given NO = 15g
Step 1 : Convert the given mass of NO to amount in mole.
Amount in mole = mass/molar mass
Molar mass of NO = Atomic mass of Nitrogen + Atomic mass of Oxygen
Molar mass of NO = 14+16 = 30g/mol
Therefore,
Amount in mole of the given NO mass = 15/30 mol
=0.5 mole
Hence,
From the given equation
6NO + 4NH3 → 5N2 + 6H2O
6 mol of NO requires 4 mol of NH3
Let 0.5 mol of NO requires x mol of NH3
x= (0.5×4)/6 mol of NH3
x= 0.33 mol of NH3
That is, 0.5 moles of NO requires 0.33 mol of NH3
At STP,
I mole of a gas = 22.4 Litres
0.33 moles of NH3 = 22.4×0.33 litres
= 7.47 litres
Therefore, 7.47 litres of NH3, at STP, is required to react with 15g of NO.
