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 L)}}[/tex]
[tex]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[/tex]
[tex]NaHCO_3+HC_2H_3O_2\rightarrow NaC_2H_3O_2+H_2O+CO_2[/tex]
According to stoichiometry:
1 mole of [tex]HC_2H_3O_2[/tex] will give = 1 mole of [tex]CO_2[/tex]
0.0208 moles of [tex]HC_2H_3O_2[/tex] will give =[tex]\frac{1}{1}\times 0.0208=0.0208 moles[/tex] of [tex]CO_2[/tex]
Mass of [tex]HC_2H_3O_2=moles\times {\text {molar mass}}=0.0208\times 60g/mol=1.25g[/tex]
Thus 1.25 g of [tex]CO_2[/tex] would be produced from the complete reaction of 25 mL of 0.833 mol/L [tex]HC_3H_3O_2[/tex] with excess [tex]NaHCO_3[/tex]
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