Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is produced from the reaction of of sulfuric acid and of sodium hydroxide, calculate the percent yield of water. Round your answer to significant figures.

This is an incomplete question, here is a complete question.

Aqueous sulfuric acid (H₂SO₄) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO₄) and liquid water (H₂O) . If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Round your answer in significant figures.

The percent yield of water is, 46.8 %

Explanation : Given,

Mass of = 72.6 g

Mass of = 77.0 g

Molar mass of = 98 g/mol

Molar mass of = 40 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 1 mole of react with 2 mole of

So, 0.741 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 1 mole of react to give 2 mole of

So, 0.741 moles of react to give moles of

Now we have to calculate the mass of

Molar mass of = 18 g/mole

Now we have to calculate the percent yield of water.

Now put all the given values in this formula, we get:

Thus, the percent yield of water is, 46.8 %

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