Answer to Question #22095 in Inorganic Chemistry for khadijah

Question #22095
If 10.00 g of magnesium is reacted with 95.75 g of copper (II) sulphate, magnesium sulphate and copper are formed.



a. Which reactant is in excess?
b. Calculate the mass of copper formed.
1
Expert's answer
2013-01-15T08:28:53-0500

Mg + CuSO4 = MgSO4 + Cu

n(Mg) = m/M

M(Mg) = 24 g/mol

n(Mg) = 10g/24g/mol = 0.417 mol

n(CuSO4) = m/M

M(CuSO4)= 64 + 32 + 16*4 = 160 g/mol

n(CnSO4) = 97.75g/160g/mol = 0.61 mol

a. Cu is in excess

n(Cu)formed = 0.417 mol

m(Cu)formed = 0.417mol*64g/mol = 26.68 g

b. The mass of copper formed = 26.68 g

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