Question

Baking soda (sodium bicarbonate) is often used to neutralize spills of acids on the benchtop of the laboratory. What mass of sodium bicarbonate would be needed to neutralize a spill consisting of 25.2 mL of 6.01 M hydrochloric acid solution?

Answer 1

Answer : The mass of $NaHCO_3$ needed is, 12.7 grams.

Explanation :

First we have to calculate the moles of HCl.

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of }HCl$

$\text{Moles of }HCl=6.01M\times 0.0252L=0.151mol$

Now we have to calculate the moles of $NaHCO_3$

The balanced neutralization reaction will be:

$HCl+NaHCO_3\rightarrow NaCl+H_2O+CO_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of $NaHCO_3$

So, 0.151 mole of HCl neutralizes by 0.151 mole of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of $NaHCO_3$ = 84 g/mole

$\text{ Mass of }NaHCO_3=(0.151moles)\times (84g/mole)=12.7g$

Therefore, the mass of $NaHCO_3$ needed is, 12.7 grams.