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The exact weight of sodium hydrogen carbonate is 4.15g using the equation you wrote above (NaHCO3 + CH3OOH ---> H2CO3 + NaCH3OO) what is the theoretical yield of this reaction? Please show work! I need to check my answer.
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Answers (1)
assuming there is the theo. amounts of reactants, then determine the moles of NaHCO3-
moles NaHCO3: 4.15 g /84.00 g/mole NaHCO3=0.0494
0.0494 moles x molec. wt. H2CO3= theo. yield of H2CO3
0.0494 moles x molec. wt. NaCH3OO= theo. wt. of sodium acetate