Question

If 28 ml of 5.8 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

Answer 1

Answer: The mass of sodium hydrogen carbonate that must be added is 27.28 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of sulfuric acid solution = 5.8 M

Volume of solution = 28 mL

Putting values in above equation, we get:

$5.8M=\frac{\text{Moles of sulfuric acid}\times 1000}{28mL}\\\\\text{Moles of sulfuric acid}=0.1624mol$

The chemical equation for the reaction of sulfuric acid and sodium hydrogen carbonate follows:

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

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of sodium hydrogen carbonate.

So, 0.1624 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.1624=0.3248mol$ of sodium hydrogen carbonate

To calculate the number of moles, we use the equation:

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

Molar mass of sodium hydrogen carbonate = 84 g/mol

Moles of sodium hydrogen carbonate = 0.3248 moles

Putting values in above equation, we get:

$0.3248mol=\frac{\text{Mass of sodium hydrogen carbonate}}{84g/mol}\\\\\text{Mass of sodium hydrogen carbonate}=(0.3248mol\times 84g/mol)=27.28g$

Hence, the mass of sodium hydrogen carbonate that must be added is 27.28 g

Answer 2

First we have to refer to the reaction between the acid and the base: 

H2SO4 + 2 NaHCO3 ---> 2 H2O + 2 CO2 + Na2SO4 

From this balanced equation we can see that for every 1 mol of acid (H2SO4), we need 2 mol of base (NaHCO3) to neutralize it. Given 28 ml of 5.8 M acid, we need to find out how many mols of acid that is: 

28mL * (1L/1000mL) * 5.8 mol/L =  0.1624 mol H2SO4

Since we need 2 mol of base per mol of acid, we need:

 2*0.1624 mol = 0.3248 mol NaHCO3 

MolarMass of NaHCO3 is 84.01 g/mol 

0.3248 mol*(84.01g/mol) = 27.29 g NaHCO3