Question

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane.

2 NH₃(g) + 3 O₂(g) + 2 CH₄(g) → 2 HCN(g) + 6 H₂O(g)

If 5.29 ✕ 10³ kg each of NH₃, O₂, and CH₄ are reacted, what mass of HCN and of H₂O will be produced, assuming 100% yield?

HCN	g
H2O	g

A solution is prepared by dissolving 14.4 g ammonium sulfate in enough water to make 113.0 mL of stock solution. A 13.20-mL sample of this stock solution is added to 52.30 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.


An 11.7-L sample of gas is determined to contain 0.40 mol N₂. At the same temperature and pressure, how many moles of gas would there be in a 10.-L sample?

Answer 1

1) We need to use the concept of limiting reagents to find out the amount of HCN and H₂O produced.

2 NH₃ (g) + 3 O₂ (g) + 2 CH₄ (g) --------> 2 HCN (g) + 6 H₂O (g)

The molar ratio is 2:3:2:2:2:6. Also, we note the molar mass of the reactants and the products as

NH₃: 17 gm/mol                                             HCN: 27 gm/mol

O₂: 32 gm/mol                                                H₂O: 18 gm/mol

CH₄: 16 gm/mol

To determine the limiting reagent, we can work with either of the products. Let us work with HCN here.

Amount of each reactant added = 5.29*10³ kg = 5.29*10⁶ gm (since 1 kg = 10³ gm)

NH₃: 5.29*10⁶ gm NH₃*(1 mole NH₃/17 gm NH₃)*(2 mole HCN/2mole NH₃)*(27 gm HCN/1 mole HCN) = 8.40*10⁶ gm HCN = 8.40*10³ HCN.

O₂: 5.29*10⁶ gm O₂*(1 mole O₂/32 gm O₂)*(2 mole HCN/3 mole O₂)*(27 gm HCN/1 mole HCN) = 2.98*10⁶ gm HCN = 2.98*10³ kg HCN.

CH₄: 5.29*10⁶ gm CH₄*(1 mole CH₄/16 gm CH₄)*(2 mole HCN/2 mole CH₄)*(27 gm HCN/1 mole HCN) = 8.93*10⁶ gm HCN = 8.93*10³ kg HCN.

The reagent which produces the smallest amount of the product is the limiting reagent. In this case, O₂ is the limiting reagent. All the O₂ will be used up and the amount of HCN produced is 2.98*10³ kg (ans).

The amount of H₂O produced = 5.29 gm O₂*(1 mole O₂/32 gm O₂)*(6 mole H₂O/3 mole O₂)*(18 gm H₂O/1 mole H₂O) = 5.95*10⁶ gm = 5.95*10³ kg (ans).

2) Molar mass of ammonium sulfate, (NH₄)₂SO₄ = 132 gm/mol.

Moles of ammonium sulfate added = 14.4 gm/(132 gm/mole) = 0.109 mole (I will round off later).

0.109 mole ammonium sulfate is dissolved in 113.0 mL = 0.113 L water. The molar concentration of ammonium sulfate = 0.109 mole/0.113 L = 0.9646 mole/L = 0.9646 M.

Now, we add 52.30 mL water to 13.20 mL stock solution. The total volume of the solution is 65.50 mL = 0.0655 L.

The molar concentration of ammonium sulfate in the final solution is

(0.0132 L)*(0.9646 mole/L)/0.0655 L = 0.194 M.

Now, we need to consider the dissociation of ammonium sulfate in water.

(NH₄)₂SO₄ (aq) <========> 2 NH₄⁺ (aq) + SO₄^2- (aq)

The molar ratio of (NH₄)₂SO₄:NH₄⁺:SO₄^2- is 1:2:1.

Therefore the molar concentration of ammonium ions = 2*(0.194 M) = 0.388 M ≈ 0.39 M (ans).

The molar concentration of sulfate ions = 1*(0.194 M) = 0.194 M ≈ 0.19 M (ans)

3) We will assume ideal behavior. Also, we assume the pressure p and the temperature T to remain constant. Let n be the moles of gas present in 10.0 L sample. Therefore, we have

p*(11.7 L) = (0.40 mole)*R*T where R = universal gas constant

===> p = (0.40 mole/11.7 L)*R*T

Again,

p*(10.0 L) = n*R*T

===> p = (n/10.0 L)*R*T

Therefore, we have,

(0.40 mole/11.7 L)*R*T = (n/10.0 L)*R*T

===> n = 0.40*10.0 mole/11.7 = 0.341 mole ≈ 0.34 mole (ans)