Answer :
Answer:
the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex] is 0.0173 moles
Explanation:
The balanced chemical equation for the reaction is represented by:
[tex]H_3C_6H_5O_{7(aq)} + 3NaHCO_{3(aq)} ------>3CO_{2(g)}+3H_2O+Na_3C_6H_5O_{7(aq)}[/tex]
From above equation; we would realize that 3 moles of [tex]NaHCO_{3(aq)}[/tex] reacts with [tex]H_3C_6H_5O_{7(aq)}[/tex] to produce 3 moles of [tex]CO_{2(aq)}[/tex]
However ; the molar mass of [tex]NaHCO_{3(aq)}[/tex] = 84 g/mol
mass given for [tex]NaHCO_{3(aq)}[/tex] = 1.45 g
therefore , we can calculate the number of moles of [tex]NaHCO_{3(aq)}[/tex] by using the expression :
number of moles of [tex]NaHCO_{3(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]
number of moles of [tex]NaHCO_{3(aq)}[/tex] = [tex]\frac{1.45}{84}[/tex]
number of moles of [tex]NaHCO_{3(aq)}[/tex] = 0.0173 mole
Since the ratio of [tex]NaHCO_{3(aq)}[/tex] to [tex]CO_{2(aq)}[/tex] is 1:1; that implies that number of moles of [tex]NaHCO_{3(aq)}[/tex] is equal to number of moles of [tex]CO_{2(aq)}[/tex] produced.
number of moles of [tex]CO_{2(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]
0.0173 = [tex]\frac{mass \ given}{ 44 \ g/mol}[/tex]
mass of [tex]CO_{2(aq)}[/tex] = 0.0173 × 44
mass of [tex]CO_{2(aq)}[/tex] = 0.7612 g
Thus; the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex] is 0.0173 moles