10.0 grams of sodium metal react with 60.0 grams of copper (ii) sulfate.
write the balanced reaction with the correct products.
identify the reaction type (single replacement, double replacement, synthesis, decomposition or combustion).
what is the limiting reactant?
what is the theoretical yield of copper metal (in grams)?
what is the percent yield if 9.75 grams of copper metal are actually produced?
( show work, the would be super !

1) 2Na + CuSO₄ → Cu + Na₂SO₄.

2) single replacement.

3) The limiting reactant is Na metal.

4) 13.80 g.

5) 70.65 %.

Explanation:

Q1: Write the balanced reaction with the correct products.

Na metal reacts with copper (II) sulfate according to the balanced reaction:

2Na + CuSO₄ → Cu + Na₂SO₄.

It is clear that 2.0 moles of Na reacts with 1.0 mole of CuSO₄ to produce 1.0 mole of Cu and 1.0 mole of Na₂SO₄.

Q2: Identify the reaction type (single replacement, double replacement, synthesis, decomposition or combustion).  

The type of the reaction is single replacement that Na replaces Cu in its salt.

Q3: What is the limiting reactant?

no. of moles of Na metal = mass/molar mass = (10.0 g)/(23.0 g/mol) = 0.4347 mol.

no. of moles of copper (II) sulfate = mass/molar mass = (60.0 g)/(159.609 g/mol) = 0.376 mol.

∴ 0.4347 mol of Na metal will react completely with 0.21735 mol of Cu (II) sulfate and its remaining will be in excess (0.376 mol - 0.21735 mol = 0.1586 mol).

So, the limiting reactant is Na metal.

Q4: What is the theoretical yield of copper metal (in grams)?

From the balanced reaction:

2.0 moles of Na produce → 1.0 mole of Cu.

∴ 0.4347 mol of Na produce → ??? mole of Cu.

∴ The no. of moles of produced Cu = (0.4347 mol)(1.0 mol)/(2.0 mol) = 0.21735 mol.

∴ The theoretical yield of Cu = no. of moles of produced Cu x atomic mass of Cu = (0.21735 mol)(63.54 g/mol) = 13.80 g.

Q5: What is the percent yield if 9.75 grams of copper metal are actually produced?

The percent yield = [actual yield/theoretical yield] x 100 = [(9.75 g)/(13.80 g)] x 100 = 70.65 %.

your answer would be “dry”