Question

What metal, and how much (g) is formed when 1.75g Zn is added to 50 mL of 0.75 M
CuSO4?​

Answer 1

1.703 g Copper

Step-by-step explanation:

We can begin by writing the balanced equation for the reaction;

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

We are given;

Mass of Zn = 1.75 g

Volume of CuSO₄ = 50 mL

Molarity of CuSO₄ = 0.75 M

Required to identify the metal formed and its mass

Step 1: Calculate the moles of Zn and CuSO₄ given

Moles of Zn

Moles = Mass ÷ Molar mass

Molar mass of Zn = 65.38 g/mol

Moles of Zn = 1.75 g ÷ 65.38 g/mol

= 0.0268 moles

Moles of CuSO₄

Moles = Molarity × Volume

Moles of CuSO₄ = 0.75 M × 0.050 L

= 0.0375 moles

Since, from the reaction 1 mole of Zn reacts with 1 mole of CuSO₄

Then, Zn is the rate limiting reaction since it has less number of moles available.

Step 2: Calculate the number of moles of Cu

1 mole of Zn reacts to form 1 mole of Cu

Thus, number of moles of Zn will be equal to the number of moles of Cu

Hence, moles of copper = 0.0268 mole

Step 3: Mass of Copper, Cu

Mass = Moles × Molar mass

Molar mass of Copper = 63.546 g/mol

Therefore;

Mass of Copper = 0.0268 moles × 63.546 g/mol

= 1.703 g

Therefore, the metal formed is copper with a mass of 1.703 g