Question

What is the percent yield of sodium hydroxide in the given reaction? The reaction was performed using 45 g NaHCO3 and 18 g NaOH were produced. NaHCO3 → NaOH + CO2

Answer 1

Answer:

The percent yield = 83.33 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $NaHCO_3$  :-

Mass of water = 45 g

Molar mass of $NaHCO_3$  = 84.007 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{45\ g}{84.007\ g/mol}$

$Moles\ of\ NaHCO_3= 0.54\ mol$

For $NaOH$  :-

Given mass = 18 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{18\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.45\ mol$

According to the given reaction:

$NaHCO_3\rightarrow NaOH+CO_2$

1 mole of $NaHCO_3$  on reaction forms 1 mole of $NaOH$

Thus,

0.54 mole of $NaHCO_3$  on reaction forms 0.54 mole of $NaOH$

Moles of $NaOH$ formed = 0.54 moles

Moles of  $NaOH$ actually formed = 0.45 moles

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 0.54 moles

Experimental yield = 0.45 moles

Applying the values in the above expression as:-

$\%\ yield =\frac{0.45}{0.54}\times 100$

$\%\ yield =83.33\ \%$

The percent yield = 83.33 %