Question

1. What is the change in pressure when 10.2 grams of molecule A (MW = 148 g/mole) is reacted with 225 mL of a 0.20 M solution of B and the resulting gas fills a volume of 500 mL vessel at 325 Kelvin? (Use the balanced equation below.)

2A(s) + B(aq) → X(g) + 2Y(aq)

a. 1.8

b. 2.4

c. 22.4

d. 186

e. 243

Answer 1

Number of moles of A = 10.2g/148g/mol = 0.0689mol

Number of moles of B = 0.20mol/L × 225×10^-3L = 0.045mol

The number of moles of A required to react completely with reagent B is 0.045×2 = 0.090mol . Hence reagent A is present in less amount so it is limiting reagent .

The number of moles of X(g) formed = moles of A ÷ 2

= 0.0689/2 = 0.03446mol

P = nRT/V = 0.03446mol×0.0821atm-L/K.mol × 325K/(500×10^-3 L)

= 1.8389 atm

The pressure change = 1.8atm  

Answer : (a) 1.8 atm