During an experiment, a student spilt 1.0 mL of concentrated H2SO4. The student neutralized the acid with 3.05 g of NaHCO3. What is the concentration of H2SO4?

18.2 M

Explanation:

M(NaHCO3) = 23 + 1+ 12+ 3*16 = 84 g/mol

3.05 g * 1 mol/84 g = (3.05/84) mol NaHCO3

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

from reaction              2 mol             1 mol

given                       (3.05/84) mol     x mol

x = (3.05/84) *1/2 = (3.05/84) /2 mol H2SO4

(3.05/84) /2 mol H2SO4 - 0.001 L

x   mol                              - 1 L

x = [(3.05/84) *1/2]/0.001 = 18 2  mol/L = 18.2 M H2SO4

the answer to your question is c.

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