Answer:
The equilibrium pressure of NO2 is 0.084 atm
Explanation:
Step 1: Data given
A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3.
Kp = 0.0118
Step 2: The balanced equation
NO( g) + SO3( g) ⇌ NO2( g) + SO2( g)
Step 3: The initial pressures
p(NO) = 0.86 atm
p(SO3) = 0.86 atm
p(NO2) = 0 atm
p(SO2) = 0 atm
Step 4: The pressure at the equilibrium
For 1 mol NO we need 1 mol SO3 to produce 1 mol NO2 and 1 mol SO2
p(NO) = 0.86 -x atm
p(SO3) = 0.86 -xatm
p(NO2) = x atm
p(SO2) = x atm
Step 5: Define Kp
Kp = ((pNO2)*(pSO2)) / ((pNO)*(pSO3))
Kp = 0.0118 = x²/(0.86 - x)²
X = 0.08427
p(NO) = 0.86 -0.08427 = 0.77573 atm
p(SO3) = 0.86 -0.08427 = 0.77573 atm
p(NO2) = 0.08427 atm
p(SO2) = 0.08427 atm
The equilibrium pressure of NO2 is 0.08427 atm ≈ 0.084 atm
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The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂