FIRST PART
Step 1
The reaction must be completed and balanced:
2NH3 (g) + 1 H2SO4 = 1 (NH4)2SO4 (aq)
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Step 2
The moles of NH3 are calculated by stoichiometry:
Information needed:
The molar mass of H2SO4 = 98 g/mol
Information provided:
The mass of H2SO4 = 1500 g
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Step 3
Procedure:
2 moles NH3 ------- 98 g H2SO4
X ------- 1500 g H2SO4
X = 1500 g H2SO4 x 2 moles NH3/ 98 g H2SO4 = 30.6 moles NH3
First part: Moles of NH3 = 30.6 moles
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SECOND PART
Step 4
NH3 gas is assumed to be ideal, therefore, it is applied:
p x V = n x R x T (1)
p = pressure = 25 atm
T = temperature (absolute) = 24 °C + 273 = 297 K
n = moles (from part 1) = 30.6 moles
R = gas constant = 0.082 atm L/mol K
V = volume = unknown
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Step 5
V is cleared from (1):
V = n x R x T/p = (30.6 moles x 0.082 atm L/mol K x 297 K)/25 atm = 29.8 L
Answer: Volume NH3 = 29.8 L
