Answer:
11.72 grams
Explanation:
Let the equilibrium concentration of BrCl be y
Initial concentration of Br2 = number of moles ÷ volume = (0.979×1000/160) ÷ 199 = 0.031 M
Initial concentration of Cl2 = (1.075×1000/71) ÷ 199 = 0.076 M
From the equation of reaction
1 mole of Br2 reacted with 1 mole of Cl2 to form 2 moles of BrCl
Therefore, equilibrium concentration of Br2 = (0.031 - 0.5y) M while that of Cl2 = (0.076 - 0.5y) M
Kp = [BrCl]^2/[Br2][Cl2]
1.1×10^-4 = y^2/(0.031 - 0.5y)(0.076 - 0.5y)
y^2/0.002356-0.0535y+0.25y^2 = 0.00011
y^2/0.00011 = 0.002356-0.0536y+0.25y^2
9090.9y^2-0.25y^2+0.0536y-0.002356 = 0
9090.65y^2+0.0535y-0.002356 = 0
The value of y must be positive and is obtained using the quadratic formula
y = [-0.0535 + sqrt(0.0535^2 - 4×9090.65×-0.002356)] ÷ 2(9090.65) = 9.2025/18181.3 = 0.00051 M
Mass of BrCl = concentration×volume×MW = 0.00051×199×115.5 = 11.72 grams
