Chemistry, 29.03.2021 21:10 rosieanneanney
Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial.
Trial 1: 0.251Mg/ 1 x 1 mole mg/ 24.305Mg x 2mol MgO/2 mol Mg x 40.304 MgO/ 1 mole MgO=
Trial 2: 0.314Mg/ 1 x 1 mole mg/ 24.305Mg x 2mol MgO/2 mol Mg x 40.304 MgO/ 1 mole MgO=
Answers: 3
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Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for ea...
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