Where's the question. I assume you want to know if the acid has been neutralized or if some remains unreacted.
..............2NaHCO3 + H2SO4 ==> Na2SO4 + 2H2O + 2CO2
mols H2SO4 = M x L = 18 x 0.0105 = 0.189
How many mols NaHCO3 will that use? That will be
0.189 mols H2SO4 x (2 mols NaHCO3/1 mol H2SO4) = 0.378
Then grams NaHCO3 = 0.378 mols x 84 g/mol = 31.75 g needed.
Initial mass NaHCO3 = 165.9 g
final mass NaHCO3 = 145.5
NaHCO3 used = 165.9 - 145.5 = 20,4 g
Conclusion: You needed 31.75 g NaHCO3, you used 20.4 g NaHCO3; therefore, some H2SO4 was left unreacted on the lab bench. How much additional NaHCO3 was needed? That 31.75 - 20.4 = ? g
Post your work if you get stuck.
To neutralize the acid in 10.5 mL of 18.0 M H2SO4 that was accidently spilled on a laboratory bench top, solid sodium bicarbonate (sodium hydrogen carbonate) was used. The container of sodium bicarbonate was known to weigh 165.9 g before this use and out of curiosity its mass was measured as 145.5 g afterward. The reaction forms sodium sulfate.
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