The mass of the Na2CO3 in the test tube after the reaction in Part A of the experiment is 0.7726 g. The theoretical yield is 0.9881 g. Find percent yield for this reaction. Write only the number, not the % symbol. (NaHCO3 = 84.01 g/mol, Na2CO3 = 105.99 g/mol) %3D %3D 2NaHCO3(s) – Na2CO3(s) + CO2(g) + H2O(g) Your Answer: Answer

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