What metal, and how much (g) is formed when 1.75g zn is added to 50 ml of 0.75 m
cuso4?

1.703 g Copper

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

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.

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

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

explanation: no but i do not usually comment