The balanced equation between the [tex] H_2SO_4 [/tex] and [tex] NaHCO_3 [/tex] is:
[tex] H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O [/tex]
Formula of molarity is:
[tex] Molarity = \frac{Moles of solute}{Volume of solution in Liters} [/tex]
[tex] Molarity = 6.0 M, Volume = 35 mL = 0.035 L [/tex]
Substituting the values,
[tex] 6 = \frac{Moles of solute}{0.035} [/tex]
[tex] Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole [/tex]
So, number of moles of [tex] H_2SO_4 [/tex] is [tex] 0.21 mole [/tex].
From the balanced equation it is clear that for 1 mole of [tex] H_2SO_4 [/tex], 2 moles of [tex] NaHCO_3 [/tex] are required.
Hence, 0.21 mole of [tex] H_2SO_4 [/tex] = [tex] 2\times 0.21 mole = 0.42 mole of NaHCO_3 [/tex]
Molar mass of [tex] NaHCO_3 [/tex] = [tex] 84.007 g/mol [/tex]
So, the mass of [tex] NaHCO_3 [/tex] = [tex] 84.007 g/mol \times 0.42 mol = 35.283 g [/tex]
