Using this balanced equation: 2 naoh + h2so4 —> h2o + na2so4

how many grams of sodium sulfate will be formed if you start with 150 grams of
sodium hydroxide and you have an excess of sulfuric acid?

!

266.325 g

Explanation:

We are given the balanced equation;

2NaOH + H₂SO₄ → H₂O + Na₂SO₄

Mass of NaOH as 150 g

We are required to determine the mass of Na₂SO₄ that will be formed.

Step 1: Determine the number of moles of NaOH

Moles = Mass ÷ molar mass

Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 150 g ÷ 40 g/mol

                          = 3.75 moles

Step 2: Determine the number of moles of sodium sulfate formedFrom the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1

Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2

                                      = 3.75 moles ÷ 2

                                     = 1.875 moles

Step 3: Determine the mass of Na₂SO₄ produced.

we know that;

Mass = Moles × Molar mass

Molar mass of Na₂SO₄ is 142.04 g/mol

Therefore;

Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol

                           = 266.325 g

Thus, the mass of sodium sulfate formed 266.325 g

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