Question

If 28 ml of 5.4 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

Answer 1

A reaction between acid and base to form water and salt is known as neutralization reaction. It is a double replacement reaction.  

The reaction between $H_2SO_4$ and $NaHCO_3$ will be:

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

The balanced reaction is:

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

Volume of $H_2SO_4$ = 28 mL (given)

Since, 1 L = 1000 mL

So, 28 mL = 0.028 L

Molarity of $H_2SO_4$ = 5.4 M (given)

Molarity = $\frac{number of moles of solute}{Volume of solution in Liters}$   -(1)

Substituting the values in equation (1):

$5.4 mol/L = \frac{number of moles of H_2SO_4}{0.028 L}$

$number of moles of H_2SO_4 = 5.4 mol/L\times0.028 L$

$number of moles of H_2SO_4 = 0.1512 mol$

From the balanced reaction between $H_2SO_4$ and $NaHCO_3$, 2 moles of $NaHCO_3$ reacts with 1 mole of $H_2SO_4$.

Molar mass of $NaHCO_3$ = 84.007 g/mol

Mass of $NaHCO_3$ needed:

$0.1512 mol H_2SO_4 \times \frac{2 mole NaHCO_3}{1 mole H_2SO_4}\times \frac{84.007 g}{1 mole NaHCO_3}$ = $25.404 g$

Hence, the required amount of $NaHCO_3$ is $25.404 g$.