Respuesta :
Answer:
The percent yield = 83.33 %
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]NaHCO_3[/tex] :-
Mass of water = 45 g
Molar mass of [tex]NaHCO_3[/tex] = 84.007 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{45\ g}{84.007\ g/mol}[/tex]
[tex]Moles\ of\ NaHCO_3= 0.54\ mol[/tex]
For [tex]NaOH[/tex] :-
Given mass = 18 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{18\ g}{39.997\ g/mol}[/tex]
[tex]Moles\ of\ NaOH= 0.45\ mol[/tex]
According to the given reaction:
[tex]NaHCO_3\rightarrow NaOH+CO_2[/tex]
1 mole of [tex]NaHCO_3[/tex] on reaction forms 1 mole of [tex]NaOH[/tex]
Thus,
0.54 mole of [tex]NaHCO_3[/tex] on reaction forms 0.54 mole of [tex]NaOH[/tex]
Moles of [tex]NaOH[/tex] formed = 0.54 moles
Moles of [tex]NaOH[/tex] actually formed = 0.45 moles
The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-
[tex]\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100[/tex]
Given , Values from the question:-
Theoretical yield = 0.54 moles
Experimental yield = 0.45 moles
Applying the values in the above expression as:-
[tex]\%\ yield =\frac{0.45}{0.54}\times 100[/tex]
[tex]\%\ yield =83.33\ \%[/tex]
The percent yield = 83.33 %
