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This means that one liter of the solution of sulfuric acid contains 0.2 gram-equivalent mass of sulfuric acid. For this acid, the equivalent mass is one half the molar mass, since each molecule of H2SO4 supplies two hydrogen atoms to neutralize alkaline materials.

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Q: When someone says 0.2 N sulfuric acid what is meant by 0.2 N?
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How many moles of excess sulfuric acid are left over after the complete reaction of 300.0 g of ammonia with 36.0 moles of sulfuric acid producing ammonium sulfate?

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 300 g of ammonia requires 300 g ÷ 34 g/mol ≈ 8.8 moles of sulphuric acid Therefore there will be 36.0 - 8.8 = 27.2 moles of sulphuric acid left.


300.0g of ammonia react with 36.0 moles of sulfuric acid to produce ammonium sulfate how many moles of excess sulfuric acid are left over after the reaction is complete?

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.

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