If 32 ml of 7.0 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

The complete balanced reaction of this neutralization reaction is:

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

Then we calculate the moles of H2SO4 that was spilled:

moles H2SO4 = 7 mole/L * 0.032 L = 0.224 mole

From the reaction, we see that 2 moles of NaHCO3 is required for every mole of H2SO4, hence:

moles NaHCO3 = 0.224 * 2 = 0.448 mole

The molar mass of NaHCO3 is 84 g/mol. Hence the mass is:

mass NaHCO3 = 0.448 * 84

mass NaHCO3 = 37.632 grams