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A reaction between 1.7 moles of zinc iodide and excess sodium carbonate yields 12.6 grams of zinc carbonate. This is the equation for the reaction: Na2CO3 + ZnI2 → 2NaI + ZnCO3. What is the percent yield of zinc carbonate? |
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Answer» A reaction between 1.7 moles of zinc iodide and excess sodium carbonate yields 12.6 grams of zinc carbonate. The EQUATION for the reaction is given as FOLLOWS: Na₂CO₃ + ZnI₂ → 2NaI + ZnCO₃. Then the percent yield of zinc carbonate can be calculated as follows; First the mole ratio of zinc iodide to zinc carbonate is 1:1 therefore 1.7 moles of zinc carbonate is theoretically produced. So theoretical yield of zinc carbonate is= Moles x relative formula MASS of zinc carbonate 1.7 x 124= 210.8 grams Hence PERCENTAGE yield= ACTUAL yield/theoretical yield x 100 = 12.6/210.8 x 100 = 6%. |
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