Question

estimate the amount of hydrochloric acid present in 500 ml of the given solution 0.05 molar sodium carbonate solution is supplied ​

Answer 1

The mass of HCl present is 1.825 grams

Explanation:

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

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

Molarity of sodium carbonate solution = 0.05 M

Volume of solution = 500 mL = 0.500 L   (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

$0.05M=\frac{\text{Moles of sodium carbonate}}{0.500L}\\\\\text{Moles of sodium carbonate}=(0.05mol/L\times 0.500L)=0.025mol$

The chemical equation for the reaction of sodium carbonate and HCl follows:

$Na_2CO_3+2HCl\rightarrow 2NaCl+CO_2+H_2O$

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.025 moles of sodium carbonate will react with = $\frac{2}{1}\times 0.025=0.05mol$ of HCl

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

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

Molar mass of HCl = 36.5 g/mol

Moles of HCl = 0.05 moles

Putting values in above equation, we get:

$0.05mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.05mol\times 36.5g/mol)=1.825g$

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