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?

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