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

The percent yield = 83.33 %

Explanation:

The formula for the calculation of moles is shown below:

For  :-

Mass of water = 45 g

Molar mass of  = 84.007 g/mol

The formula for the calculation of moles is shown below:

Thus,

For  :-

Given mass = 18 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of  on reaction forms 1 mole of

Thus,

0.54 mole of  on reaction forms 0.54 mole of

Moles of formed = 0.54 moles

Moles of   actually formed = 0.45 moles

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

Given , Values from the question:-

Theoretical yield = 0.54 moles

Experimental yield = 0.45 moles

Applying the values in the above expression as:-

The percent yield = 83.33 %

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