Respuesta :
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
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:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of sulfuric acid solution = 5.8 M
Volume of solution = 28 mL
Putting values in above equation, we get:
[tex]5.8M=\frac{\text{Moles of sulfuric acid}\times 1000}{28mL}\\\\\text{Moles of sulfuric acid}=0.1624mol[/tex]
The chemical equation for the reaction of sulfuric acid and sodium hydrogen carbonate follows:
[tex]H_2SO_4(aq.)+2NaHCO_3(aq.)\rightarrow Na_2SO_4(aq.)+2CO_2(g)+H_2O(l)[/tex]
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 = [tex]\frac{2}{1}\times 0.1624=0.3248mol[/tex] of sodium hydrogen carbonate
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of sodium hydrogen carbonate = 84 g/mol
Moles of sodium hydrogen carbonate = 0.3248 moles
Putting values in above equation, we get:
[tex]0.3248mol=\frac{\text{Mass of sodium hydrogen carbonate}}{84g/mol}\\\\\text{Mass of sodium hydrogen carbonate}=(0.3248mol\times 84g/mol)=27.28g[/tex]
Hence, the mass of sodium hydrogen carbonate that must be added is 27.28 g
