Question

22. A flask containing 450 mL of 0.50 M H2SO4 was accidentally knocked to the floor. How many grams of NaHCO, do you need to put on the spill to neutralize the acid according to the following equation: H2SO4(aq)+2 NaHCOs(aq) Na,SO(aq) +2 H20()+2 CO2(g) D) 38 g A) 2.3 g B) 9.5 g C) 19 g

Answer 1

Answer: The correct answer is Option D.

Explanation:

To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of sulfuric acid = 450mL = 0.45 L      (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.5 moles/ L

Putting values in above equation, we get:

$0.5mol/L=\frac{\text{Moles of sulfuric acid}}{0.45L}\\\\\text{Moles of sulfuric acid}=0.225mol$

For the given chemical reaction:

$H_2SO_4(aq.)+2NaHCO_3(aq.)\rightarrow Na_2SO_4(aq.)+2H_2O(l)+2CO_2(g)$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of sodium bicarbonate.

So, 0.225 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.225=0.45mol$ of sodium bicarbonate

To calculate the mass of sodium bicarbonate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sodium bicarbonate = 0.45 moles

Molar mass of sodium bicarbonate = 84.007 g/mol

Putting values in above equation, we get:

$0.45mol=\frac{\text{Mass of sodium bicarbonate}}{84.007g/mol}\\\\\text{Mass of sodium bicarbonate}=38g$

Hence, the correct answer is Option D.