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The reaction of citric acid and sodium bicarbonate is written as: H3C6H5O7(aq) + 3NaHCO3(aq) --> Na3C6H5O7(aq) + 3H2O(l) + 3CO2(g) 1.) Mass of the cup,water and the stirrer = 6g 22.32.) Mass of sodium bicarbonate = 1.07g 3.) Mass of citric acid =0.76g Calculate the theoretical yield of carbon dioxide from 1.07g of sodium bicarbonate. And calculate the percentage yield in the plastic cup

Explanation / Answer

Given mass of NaHCO3 = 1.07g

no. of moles of NaHCO3 present = mass/ molar mass = 1.07/84 = 0.012moles

3 moles NaHCO3 produces 3 moles CO2.

So, no of moles of CO2 produced = no. of moles of NaHCO3 = 0.012moles

Mass of CO2 to be present = no of moles x molar mass = 0.012 x 44 = 0.56g = theoretical yield of CO2.

For, percentage yield, I need the mass of cup and stirrer separately so that I can calculate mass of water and cO2 present.

For %yield = experimental yield/ theoretical yield x 100

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