What volume of ammonia gas (NH3), measured at 600 mm Hg and 55 °C, is requirec to produce 5.00 g of ammonium sulfate according to the following reaction? 2 NH3 (g) + H2S04 (aq) - (NH4)2SO4 (s)

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What volume of ammonia gas (NH3), measured at 600 mm Hg and 55 °C, is requirec
to produce 5.00 g of ammonium sulfate according to the following reaction?
2 NH3 (g) + H2S04 (aq) → (NH4)2SO4 (s)
3.76 L
O1.50 L
2.58 L
0.43 L
Transcribed Image Text:What volume of ammonia gas (NH3), measured at 600 mm Hg and 55 °C, is requirec to produce 5.00 g of ammonium sulfate according to the following reaction? 2 NH3 (g) + H2S04 (aq) → (NH4)2SO4 (s) 3.76 L O1.50 L 2.58 L 0.43 L
Expert Solution
Step 1

Given that,

Pressure = 600 mmHg , Temperature = 55⁰C = 328.15⁰C

Since, 760 mmHg = 1 atm

Hence,  

Pressure (600 mmHg) = 1 atm760 mm Hg× 600 mmHg = 0.789 atm

Mass of product (NH4)2SO4 = 5.00 g,  molar mass = 132.14 g/mol

Hence, 

moles of product = massMolar mass= 5.00 g132.14 g/mol =0.0378 mol

 

The reaction is given as:

2NH3 (g) + H2SO4(aq)→ (NH4)2SO4 (s)

Here, form reaction stoichiometry 2 mol NH3  produces 1 mol of  ammonium sulfate.

Hence, the  moles of ammonia gas required =

moles of NH3= 2 mol NH31 mol (NH4)2SO4   ×0.0378 mol(NH4)2SO4= 0.0756 mol NH3

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