Answer to Question #85265 in General Chemistry for Caleb

Question

If 376 mL of 5.52 magnesium sulfide is reacted with excess aluminum chloride according to the following balance equation
3MgS + 2AlCL3— 3MgCl2 + Al2S3

Answer 1

3MgS + 2AlCl₃ -> 3MgCl₂ + Al₂S₃

n=m/M

M(MgS) = 24.31+32.07 = 56.38 g/mol

n(MgS) = 5.52 g/ 56.38 g/mol = 0.0979 mol

According to equation 3 mol of MgS give 1 mole of Al₂S₃

We have 0.0979 mol of MgS which give x mol of Al₂S₃

Solve the proportion:

3/0.0979 = 1/x

x = 0.0326

n(Al₂S₃)= 0.0326 mol

m = M*n

M(Al₂S₃) = 26.98*2+32.07*3 = 150.17 g/mol

m(Al₂S₃) = 0.0326 mol *150.17 g/mol = 4.90 g

Answer: 4.90 g