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
