Question

If you have 35.0g of Na and 100.0g of Cl2, what is the maximum amount of NaCl that you can produce?

Answer 1

Answer:

89.0 g.

Explanation:

• The balanced chemical equation between CuO and HCl is:

2Na + Cl₂ → 2NaCl,

It is clear that 2 mol of Na react with 1 mol of Cl₂ to produce 2 mol of NaCl.

• Firstly, we need to calculate the no. of moles of (35.0 g of Na) and (100.0 g Cl₂):

no. of moles of Na = mass/molar mass = (35.0 g)/(22.989 g/mol) = 1.522 mol.

no. of moles of Cl₂ = mass/molar mass = (100.0 g)/(70.906 g/mol) = 1.41 mol.

From the balanced equation: Na reacts with Cl₂ with (2: 1) molar ratio.

∴ 1.522 mol of Na "limiting reactant" reacts completely with 0.761 mol of Cl₂ "excess reactant".

• To calculate the mass of the produced salt (CuCl₂):

2 mol of Na produces → 2 mol of NaCl, from stichiometry.

∴ 1.522 mol of Na produces → 1.522 mol of NaCl.

∴ The mass of the produced salt (NaCl) = (no. of moles)(molar mass) = (1.522 mol)(58.44 g/mol) = 88.95 g ≅ 89.0 g.