If 2.00 g of Zn is allowed to react with 2.00 g of CuSO4, how many grams of

Zn will remain after the reaction is complete? CuSO4 (aq) + Zn (s) - > ZnSO4 (aq) + Cu (s)

1.1445g

Explanation:

Mass of Zn = 2.0g

Mass of CuSO4 = 2.0g

Molar mass of Zn = 65.4g/mol

Molar mass of CuSO4 = 159.5g/mol

No. Of moles = mass / molar mass

No. Of moles of Zn = 2.0/65.4 = 0.030moles

No. Of moles of CuSO4 = 2/159.5 = 0.0125 moles

Equation of reaction

CuSO4 (aq) + Zn (s) - > ZnSO4 (aq) + Cu (s)

Stoichiometric molar ratio of CuSo4 : Zn = 1:1

Actual molar ration of CuSO4 : Zn = ?

0.0125 / 0.030 = 1 / 2.4

Zn is present in excess in the reaction, while CuSO4 is the limiting reagent

1 mol of CuSO4 = 1 mol of Zn

0.0125 moles of CuSO4 = 0.0125 moles of Zn

No. Of moles left after reaction = 0.030 - 0.0125 = 0.0175 moles.

Mass = no. Of moles * molar mass = 0.0175 * 65.4 = 1.1445g

The mass of Zn left after the reaction is 1.1445g.