Answer:
[tex]m=1.325gNa_2CO_3[/tex]
Explanation:
Hello,
In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:
[tex]eq-g=0.5eq-g/L*0.050L=0.025eq-g[/tex]
Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:
[tex]mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3[/tex]
Finally, the mass is computed by using its molar mass (106 g/mol)
[tex]m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3[/tex]
Regards.
