Final answer:
To determine the mass of sodium sulfate produced from 60.1 g of NaOH, we first convert the mass of NaOH to moles, use the 2:1 mole ratio from the balanced equation to calculate the moles of Na₂SO₄ produced, and then convert the moles of Na₂SO₄ to grams to get our final answer of approximately 106.6775 g Na₂SO₄.
Step-by-step explanation:
The question is asking how many grams of sodium sulfate, Na₂SO₄, can be produced from 60.1 g of sodium hydroxide, NaOH, in a chemical reaction with sulfuric acid, H₂SO₄. First, we need to convert the mass of NaOH to moles using its molar mass. The molar mass of NaOH is approximately 40 g/mol.
60.1 g NaOH × (1 mol NaOH / 40 g NaOH) = 1.5025 mol NaOH
According to the balanced equation 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O, it takes 2 moles of NaOH to produce 1 mole of Na₂SO₄. Therefore, the amount of NaOH we have (1.5025 mol) will produce half as many moles of Na₂SO₄ since the ratio is 2:1.
1.5025 mol NaOH / 2 = 0.75125 mol Na₂SO₄
To find the mass of Na₂SO₄ produced, multiply the number of moles by the molar mass of Na₂SO₄, which is approximately 142 g/mol.
0.75125 mol Na₂SO₄ × (142 g Na₂SO₄ / 1 mol Na₂SO₄) = 106.6775 g Na₂SO₄.
Therefore, 60.1 g of NaOH can produce 106.6775 g of Na₂SO₄.