Question

N THE FOLLOWING INFORMATION: 3 Cuso, (aq) → 3 Cu (s) + A12(SO4)3 (aq) GIVE 6.75 grams of Aluminum Density of aluminum- 2.70 g/mL Tolarity of the CuSO, solution- 0.345 M CALCULATE: A. Volume of the Al (s) sample Mass in grams of copper produced C. Volume in mL of the CuSO4 (aq) solution used D. Mass in grams of the product Al(SO)s produced

Answer 1

A)
volume of Al = (mass of Al)/(density of Al)
= (6.75/2.7) mL
= 2.5 mL
B)
number of mole Al = (mass of Al)/(molar mass of Al)
molar mass of Al = 27 g/mol
number of mole Al = 6.75/27
= 0.25 mole

according to reaction taking place
2 mol of Al give 3 mol of Cu
1 mol of Al give 3/2 mol of Cu
0.25 mol of Al give (3/2)*0.25 mol of Cu
0.25 mol of Al give 0.375 mol of Cu
number of mole of Cu = 0.375 mol
mass of Cu = (number of mole of Cu)*(molar mass of Cu)
molar mass of Cu = 63.5 g/mol
mass of Cu = (0.375*63.5) g
= 23.8 g
C)
2 mol of Al required 3 mol CuSO4
1 mol of Al required 3/2 mol CuSO4
0.25 mol of Al required (3/2)*0.25 mol CuSO4
0.25 mol of Al required 0.375 mol CuSO4
number of mole of CuSO4 required = 0.375 mole
(molarity of CuSO4)*(volume in L) = 0.375
0.345*(volume in L) = 0.375
(volume in L) = 1.087 L
volume of CuSO4 = 1087 mL
D)
2 mole of Al give 1 mole of Al2(SO4)3
1 mole of Al give 1/2 mole of Al2(SO4)3
0.25 mole of Al give (1/2)*0.25 mole of Al2(SO4)3
0.25 mole of Al give 0.125 mole of Al2(SO4)3
number of mole of Al2(SO4)3 = 0.125
mass of Al2(SO4)3 = (number of mole)*(molar mass of Al2(SO4)3)
molar mass of Al2(SO4)3 = 342.2 g/mol
mass of Al2(SO4)3 = (0.125*342.2) g
= 42.8 g