Question

Suppose 6.87 g of sulfuric acid is mixed with 9.7 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

9.96g of Na₂SO₄ is the maximum mass that could be produced

Step-by-step explanation:

We must determine the reaction:

Reactants: H₂SO₄, NaOH

Products: H₂O, Na₂SO₄

The equation is: H₂SO₄ (aq) + 2NaOH(aq) → 2H₂O (l) + Na₂SO₄(aq)

We have the mass of the reactants. We need to convert them to moles, in order to define the limiting reactant

6.87 g / 98 g/mol = 0.0701 moles of acid

9.7 g / 40 g/mol = 0.242 moles of base

Limiting reactant is the acid. Let's verify

2 moles of NaOH can react with 1 mol of acid

Therefore 0.242 moles of NaOH must react with (0.242 . 1) / 2 = 0.121 moles

We do not have enough acid.

Ratio with the salt is 1:1. 1 mol of acid produces 1 mol of salt

Therefore 0.0701 moles of acid will produce 0.0701 moles of salt

We convert the moles to mass → 0.0701 mol . 142.06 g / 1 mol = 9.96g