# Oh, no! You just spilled 85.00

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid onyour lab bench and need to clean it up immediately! Right next toyou there is a small jar labeled “15.00 g sodium bicarbonate”. Willthis be enough sodium bicarbonate to neutralize the spilledsulfuric acid? Show your work and state all reasoning.

First you need a balanced equation that describes the reactionbetween sulfuric acid and sodium bicarbonate:H2SO4(aq) + 2NaHCO3(s) –> Na2SO4(aq) + 2H2O(l) + 2CO2(g)Next you need to calculate how many moles of H2SO4 are present in85.00 mL of 1.500 M sulfuric acid.Molarity = moles of solute / liters of solution1.500 M = n / 0.08500 Ln = 0.1275 mol H2SO4Now set up and solve a stoichiometric conversion from moles ofH2SO4 to grams of NaHCO3. For this you’ll need to know the molarmass of NaHCO3.1 x Na = 1 x 22.99 = 22.991 x H = 1 x 1.01 = 1.011 x C = 1 x 12.01 = 12.013 x O = 3 x 16.00 = 48.00Total = 84.01 g/mol0.1275 mol H2SO4 x (2 mol NaHCO3 / 1 mol H2SO4) x (84.01 g NaHCO3 /1 mol NaHCO3)= 21.42 g NaHCO3So unfortunately, 15.00 grams of sodium bicarbonate will *not* besufficient to completely neutralize the acid. You would need anadditional 6.42 grams to complete the task.

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