Answer:
1) 2Na + CuSO₄ → Cu + Na₂SO₄.
2) single replacement.
3) The limiting reactant is Na metal.
4) 13.80 g.
5) 70.65 %.
Step-by-step 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?
- To determine the limiting reactant, we should calculate the no. of moles of the two reactants:
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.
- From the balanced reaction; Na reacts with Cu (II) sulfate with (2:1) ratio.
∴ 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)?
- To determine the theoretical yield of Cu, we should calculate the no. of moles of produced Cu.
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 %.