If 28 ml of 5.8 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

Question

Kiera Mccormick

Chemistry

6 April, 17:57

If 28 ml of 5.8 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

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Mercedes · Answer 1

First we have to refer to the reaction between the acid and the base:

H2SO4 + 2 NaHCO3 - - - > 2 H2O + 2 CO2 + Na2SO4

From this balanced equation we can see that for every 1 mol of acid (H2SO4), we need 2 mol of base (NaHCO3) to neutralize it. Given 28 ml of 5.8 M acid, we need to find out how many mols of acid that is:

28mL * (1L/1000mL) * 5.8 mol/L = 0.1624 mol H2SO4

Since we need 2 mol of base per mol of acid, we need:

2*0.1624 mol = 0.3248 mol NaHCO3

MolarMass of NaHCO3 is 84.01 g/mol

0.3248 mol * (84.01g/mol) = 27.29 g NaHCO3