Question

Consider the following reaction: NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2 How many g of CO2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC3H3O2 with excess NaHCO3 ?

Answer 1

1.25 g of $CO_2$ would be produced from the complete reaction of 25 mL of 0.833 mol/L $HC_3H_3O_2$ with excess $NaHCO_3$

Explanation:

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

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

$0.833M=\frac{\text{Moles of} HC_3H_3O_2\times 1000}{25ml}\\\\\text{Moles of} HC_3H_3O_2 =\frac{0.833mol/L\times 25}{1000}=0.0208mol$

$NaHCO_3+HC_2H_3O_2\rightarrow NaC_2H_3O_2+H_2O+CO_2$

According to stoichiometry:

1 mole of $HC_2H_3O_2$ will give = 1 mole of $CO_2$

0.0208 moles of $HC_2H_3O_2$ will give =$\frac{1}{1}\times 0.0208=0.0208 moles$ of $CO_2$

Mass of $HC_2H_3O_2=moles\times {\text {molar mass}}=0.0208\times 60g/mol=1.25g$

Thus 1.25 g of $CO_2$ would be produced from the complete reaction of 25 mL of 0.833 mol/L $HC_3H_3O_2$ with excess $NaHCO_3$

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