Question

If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid.

Answer 1

The balanced equation between the $H_2SO_4$ and $NaHCO_3$ is:

$H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O$

Formula of molarity is:

$Molarity = \frac{Moles of solute}{Volume of solution in Liters}$

$Molarity = 6.0 M, Volume = 35 mL = 0.035 L$

Substituting the values,

$6 = \frac{Moles of solute}{0.035}$

$Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole$

So, number of moles of $H_2SO_4$ is $0.21 mole$.

From the balanced equation it is clear that for 1 mole of $H_2SO_4$, 2 moles of $NaHCO_3$ are required.

Hence, 0.21 mole of $H_2SO_4$  = $2\times 0.21 mole = 0.42 mole of NaHCO_3$

Molar mass of $NaHCO_3$ = $84.007 g/mol$

So, the mass of $NaHCO_3$ = $84.007 g/mol \times 0.42 mol = 35.283 g$