How many grams of Cr would be needed to replace the Cu in 100 mL of a 0.75M solution of Cu(II)sulfate

Question

How many grams of Cr would be needed to replace the Cu in 100 mL of a 0.75M solution of Cu(II)sulfate

Answer 1

100 ml x 0.75 mol/ml = 0.075 mol

0.075 mol x 2/3 = 0.05 mol Cr

0.05mol Cr x 51.995g/mol = 2.6g Cr

Answer 2

2Cr + 3CuSO4 ===> Cr2(SO4)3 + 3Cu

Atomic weight: Cr=52

Let the CuSO4 solution be called S.

100mLS x 0.75molCuSO4/1000mLS x 2molCr/3molCuSO4 x 52gCr/1molCr = 2.6g Cr

Answer 3

2 Cr3+ +3 CuSO4 = Cr2(SO4)3 +3 Cu

Moles Cu2+ = moles SO42- =  0.100 L x 0.75 M = 0.075

Moles Cr2(SO4)3 = 0.075 / 3 = 0.025

Moles Cr3+ = 2 x 0.025 = 0.050

Mass Cr3+ = 0.050 mol x 52 g/mol =2.6 g