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
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
