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3 years ago

How many millimeters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3

2 answers:

8 0

So we need 0.276 moles of HCl to react. Your concentration is given in moles/liter so 0.276/1.58 = 0.174 liters needed or 174 milliliters

Tju [1.3M]3 years ago

7 0

Answer: The volume of HCl needed is 177.2 mL

Explanation:

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

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

Given mass of sodium hydrogen carbonate = 23.2 g

Molar mass of sodium hydrogen carbonate = 84 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium hydrogen carbonate}=\frac{23.2g}{84g/mol}=0.28mol$

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

$HCl(aq.)+NaHCO_3(aq.)\rightarrow NaCl(aq.)+CO_2(g)+H_2O(l)$

By Stoichiometry of the reaction:

1 mole of sodium hydrogen carbonate reacts with 1 mole of hydrochloric acid

So, 0.28 moles of sodium hydrogen carbonate will react with = $\frac{1}{1}\times 0.28=0.28mol$ of hydrochloric acid

To calculate the volume 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 hydrochloric acid = 1.58 M

Moles of hydrochloric acid = 0.28 mol

Putting values in above equation, we get:

$1.58M=\frac{0.28\times 1000}{\text{Volume of hydrochloric acid}}\\\\\text{Volume of hydrochloric acid}=177.2mL$

Hence, the volume of HCl needed is 177.2 mL

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