Question

H2SO4 + 2 NaOH → Na2SO4 + 2H2O Using the balanced chemical reaction above, determine how many grams of sodium sulfate will be formed from a solution that contains 45.00 g of sulfuric acid in an excess of sodium hydroxide.

What is the molar ratio between sodium sulfate and sulfuric acid?

Use the moles of sulfuric acid calculated above and the molar ratio from the balanced chemical reaction between the moles of sodium sulfate produced divided by the moles of sulfuric acid used to determine the number of moles of sodium sulfate that would be produced from the moles of sulfuric acid calculated in step (b).

What is the molar mass of sodium sulfate?

From the moles of sodium sulfate calculated in step (d), determine the grams of sodium sulfate that will be produced?

Answer 1

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O

98.079 g of sulfuric acid produces 142.04 g of sodium sulfate

So, 45.00 g of sulfuric acid will produce 142.04 g x( 45.00 g /98.079 g) = 65.17 g

The molar ratio between sodium sulfate and sulfuric acid is 1:1

Molar mass of sulfuric acid = 98.079 g/mol

Moles of sulfuric acid = (45.00 g of sulfuric acid)/ (98.079 g/mol of sulfuric acid)

Moles of sulfuric acid = 0.4588 moles

Sodium hydroxide is in excess, so sulfuric acid is limiting reagent.

1 mole of sulfuric acid produces 1 mole of Sodium sulfate

So, 0.4588 moles of sulfuric acid produces 0.4588 moles of Sodium sulfate

Convert 0.04588 moles of Sodium sulfate to gram of sodium sulfate by multiplying with its molar mass

Sodium sulfate has a molar mass of 142.04 g mol−1

Grams of sodium sulfate = 0.4588 moles of Sodium sulfate x142.04 g mol−1

Grams of sodium sulfate = 65.17 g