Question

What volume of ammonia gas, measured at 547.9 mmHg and 27.6oC, is required to produce 8.98 g of ammonium sulfate according to the following balanced chemical equation? 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s)

Answer 1

Answer:

4.66 L of ammonia gas will be produced

Explanation:

Step 1: Data given

The pressure of ammonia gas = 547.9 mmHg = 0.72092116 atm

Temperature = 27.6 °C = 300.75 K

Mass of ammonium sulfate produced = 8.98 gramms

Molar mass of ammonium sulfate = 132.14 g/mol

Step 2: The balanced equation

2NH3(g) + H2SO4(aq) → (NH4)2SO4(s)

Step 3: Calculate moles (NH4)2SO4

Moles (NH4)2SO4 = mass (NH4)2SO4 / molar mass

Moles (NH4)2SO4 = 8.98 grams / 132.14 g/mol

Moles (NH4)2SO4 = 0.0680 moles

Step 4: Calculate moles NH3

For 1 mol (NH4)2SO4 we need 2 moles NH3

For 0.0680 moles (NH4)2SO4 we need 2*0.0680 = 0.136 moles NH3

Step 5: Calculate volume NH4

p*V=n*R*T

V = (n*R*T)/p

⇒with V = the volume of NH3 = TO BE DETERMINED

⇒with n = the number of moles NH3 = 0.136 moles NH3

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 300.75 K

⇒with p = the pressure of the gas = 0.72092116 atm

V = (0.136 * 0.08206 * 300.75) / 0.72092116

V = 4.66 L

4.66 L of ammonia gas will be produced

Answer 2

Answer:

4.65 L of NH₃ is required for the reaction

Explanation:

2NH₃(g)  +  H₂SO₄(aq)  → (NH₄)₂SO₄(s)

We determine the ammonium sulfate's moles that have been formed.

8.98 g . 1mol / 132.06 g = 0.068 moles

Now, we propose this rule of three:

1 mol of ammonium sulfate can be produced by 2 moles of ammonia

Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.

Firstly we do unit's conversions:

27.6°C +273 =  300.6 K

547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm

V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm

V = 4.65 L