# What mass of sodium sulphate is produced when 25.0 mL of 0.100 M sulphuric acid is added to 20.0 mL of 0.15 M sodium hydroxide solution?

## Expert Answers First write a balanced chemical equation showing the neutralization reaction between sulfuric acid and sodium hydroxide.

H2SO4 + 2 NaOH -->  Na2SO4    + 2 HOH

this tells you the ratio of acid to base (1:2), and the ratio of either acid or base to amount of sodium sulfate produced.  (You get one mole of sodium sulfate for every mole of sulfuric acid and you get one mole of sodium sulfate for every two moles of sodium hydroxide)

Second: determine the moles of each reactant:

moles of sulfuric acid = L of sulfuric acid * M of sulfuric acid

moles of H2SO4 = 0.025 L * 0.1 M = 0.0025 moles

moles of sodium hydroxide = L of sodium hydroxide * M of NaOH

moles of NaOH = 0.20 * 0.15 = 0.0030 moles of NaOH

Third:

Compare the moles of each reactant to see if you have twice as much NaOH as you have of sulfuric acid. As you can see, you do not have enough NaOH for the amount of sulfuric acid. So the NaOH limits the amount of sodium sulfate you can make.

From step 1, you know that you get one mole of sodium sulfate for every two moles of NaOH, so if you have 0.0030 moles of NaOH, you will get 0.0015 moles of sodium sulfate.

Four:

Find the mass of sodium sulfate produced.  Calculate the formula mass of sodium sulfate and multiply by the # of moles and you have your final answer.

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