Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 2.9 g of sulfuric acid is mixed with 3.53 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

1.07 g of water.

Explanation:

A reaction between an acid and a base makes water and a salt as product.

Our reaction is:

H₂SO₄ +  2NaOH  →  Na₂SO₄  +  2H₂O

Reactants are the acid and the base. Which is the limiting?

2.9 g . 1mol /98 g =  0.0296 moles of acid

3.53 g . 1mol / 40 g = 0.088 moles of base

2 moles of base react to 1 mol of acid

0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid

And we only have 0.0296, sulfuric acid is the limiting

Ratio is 1:2. 1 mol of acid can produce 2 moles of water.

Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.

We convert moles to mass:

0.0592 mol . 18g /mol = 1.07 g

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