Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 48.1 g of sulfuric acid is mixed with 51. g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to significant digits.

69.7 g of Na₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

Next, we shall determine the masses of H₂SO₄ and NaOH that reacted and the mass of Na₂SO₄ produced from the balanced equation. This can be obtained as shown below:

Molar mass of H₂SO₄ = (2×1) + 32 + (4×16)

= 2 + 32 + 64

= 98 g/mol

Mass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mass of NaOH from the balanced equation = 2 × 40 = 80 g

Molar mass of Na₂SO₄ = (2×23) + 32 + (4×16)

= 46 + 32 + 64

= 142 g/mol

Mass of Na₂SO₄ from the balanced equation = 1 × 142 = 142 g

SUMMARY:

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH to produce 142 g of Na₂SO₄.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH.

Therefore, 48.1 g of H₂SO₄ will react with = (48.1 × 80)/98 = 39.3 g of NaOH.

From the calculation made above, we can see that only 39.3 g of NaOH out of 51 g given was needed to react completely with 48.1 g of H₂SO₄. Therefore, H₂SO₄ is the limiting reactant and NaOH is the excess reactant.

Finally, we shall determine the maximum mass of Na₂SO₄ produced from the reaction. This can be obtained by using the limiting reactant as illustrated below:

NOTE: H₂SO₄ is the limiting

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄.

Therefore, 48.1 g of H₂SO₄ will react to produce = (48.1 × 142)/98 = 69.7 g of Na₂SO₄.

Thus, 69.7 g of Na₂SO₄ were obtained from the reaction.

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