Suppose you spill 20.0 mL of 0.150 M sulfuric acid on the table. You decide to neutralize it with .275 grams of sodium hydroxide.

Question

A) Find the limiting reactant: Was the mass completely neutralized?

B) Determine the mass of sodium sulfate that would be recovered from the reaction?

Answer 1

This is a neutralization that proceeds according to: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

From this balanced chemical equation, we see that for every one mole of sulfuric acid, we need two moles of sodium hydroxide for a complete neutralization. Let's calculate moles of acid present.

0.150 M means we have 0.150 moles for every liter, or 1,000 mL of acid, present, so 0.150 moles/1000 mL * 20.0 mL = 0.00300 moles sulfuric acid, rounding to three significant figures.

Let's do the same for sodium hydroxide using the molar mass as a conversion factor:

0.275 g NaOH * 1 mole NaOH/40.0 g NaOH = 0.006875 moles = 0.00688 moles, rounding to three significant figures.

Now, let's see how much NaOH is required to neutralize the acid, using the balance chemical equation:

0.00300 moles acid *(2 mol NaOH/1 mole acid) = 0.00600 moles NaOH.

Since we have more that 0.00600 moles NaOH, the acid would be completely neutralized with 0.00088 moles NaOH left over.

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