Question

This balanced chemical equation represents a chemical reaction: 6no + 4nh3 → 5n2 + 6h2o what volume of nh3 gas, at standard temperature and pressure (stp), is required to react with 15.0 g of no?

Answer 1

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.

Answer 2

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.