Part B: Limiting Reactant and Excess Reactant Reaction Run 2 Measured Mass of Na

Question

Part B: Limiting Reactant and Excess Reactant Reaction Run 2 Measured Mass of NaHCO3 = 1.98 g, Measured Mass of H3C6H5O7 = 0.75 g, Measured Mass of CO2 = 0.46 g. Determine the limiting reactant in the reaction run 2. Describe your reasoning. (10 points)

Calculate the amount (in g) of carbon dioxide from the amount of the limiting reactant that you used in the reaction run 2. The amount that you calculate is called the theoretical yield. Show all your work. (10 points)

The measured mass of carbon dioxide in the table is the actual yield of carbon dioxide for the reaction run 2. Calculate the percent yield of carbon dioxide in the reaction run 2. Show all your work. (10 points)

Explanation / Answer

Solution: 3NaHCO3+ H3C6H5O7------->3CO2+3H2O+Na3C6H5O7

3moles of NaHCO3 required 1 mol of H3C6H5O7

1.98/84 moles of NaHCO3 required = 0.0078 moles

So the limiting reagent is NaHCO3.

3moles NaHCO3 forms 3 moles of CO2

So, 0.023 moles of NaHCO3 will form 0.023 moles of CO2

%yield =Actual yield×100/Theoretical yield

=0.46×100/0.023×44= 45.45%

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