Chemical Equations

How many moles of excess sulfuric acid are left over after the complete reaction of 500.0 g of ammonia with 51.0 moles of sulfuric acid producing ammonium sulfate?

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August 25, 2016 7:36AM

The equation for the reaction is:

H{2}SO{4} + 2NH{3} → (NH{4}){2}SO{4}

[Numbers in braces are supposed to be subscripts, but I can't do them here.]

This says that 1 mole of sulphuric acid reacts with 2 moles of nitrogen to create 1 mole of ammonium sulphate.

1 mole of a substance weighs the same as its atomic weight in grams.

1 mole of NH{3} weighs 14 + 3 × 1 = 17 g

Thus 2 × 17 g = 34 g of ammonia react with 1 mole of sulphuric acid.

To react with 500 g of ammonia requires 500 g ÷ 34 g/mol ≈ 14.7 moles of sulphuric acid

Therefore there will be 51.0 - 14.7 = 36.3 moles of sulphuric acid left.