Question

Students working in lab accidentally spilled 14 L of 3.0 M H2SO4 solution. They find a large container of acid neutralizer that contains baking soda, NaHCO3. How many grams of baking soda will be needed to neutralize the sulfuric acid spill? Do not include a unit in your answer or it will be counted wrong. Use correct significant figures.

Answer 1

First we have to balance the chemical reaction equation

2 NaHCO₃ (s) + H₂SO₄ (aq) --> Na₂SO₄ (aq) + 2 H₂O(l) + 2CO₂ (g)

according to above balanced equation we find 2 moles of baking soda requires 1 mole H₂SO₄ hence we have to calculate how many moles are present in 14 L of 3.0 M H2SO4 solution.

moles H2SO4 = 14 L x 3.0 M = 42 mol, this represent we need twice of this amount of NaHCO3

moles NaHCO3 required = 2 x 42 mol = 84 moles

convert moles NaHCO3 to mass NaHCO3 by multiplying with molar mass of NaHCO3 (molar mass of NaHCO3 = 84.007 g/mol)

grams of baking soda will be needed to neutralize the sulfuric acid spill = 84 mol x 84.007 g/mol = 7056.588 grams

Final answer without unit and with significant figures = 7.1 x 10³ (OR) 7.1E3

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