Question

If 1.80 g Zn is allowed to react with 1.56 g of CuSO4 according to equation CuSO4(aq)+Zn(s)→ZnSO4(aq)+Cu(s), how many grams of Zn will remain after the reaction is complete

Answer 1

Number of moles of Zn = 1.80 g / 65.38 g/mol = 0.0275 mole

Number of moles of CuSO4 = 1.56 g / 159.609 g/mol = 0.00977 mole

From the balanced equation we can say that

1 mole of Zn requires 1 mole of CuSO4 so

0.0275 mole of Zn will require

= 0.0275 mole of Zn *(1 mole of CuSO4 / 1 mole of Zn)

= 0.0275 mole of CuSO4

But we have 0.00977 mole of CuSO4 which is in short so CuSO4 is limiting reactant and Zn is an excess reactant

From the balanced equation we can say that

1 mole of CuSO4 requires 1 mole of Zn so

0.00977 mole of CuSO4 will require

= 0.00977 mole of CuSO4 *(1 mole of Zn / 1 mole of CuSO4)

= 0.00977 mole of Zn

The number of moles of Zn remain after completion of reaction = 0.0275 - 0.00977 = 0.0177 mole

mass of 1 mole of Zn = 65.38 g so

the mass of 0.0177 mole of Zn = 1.16 g

Therefore, the mass of Zn remain after completion of reaction = 1.16 g