Vivi M.

asked • 04/01/21

4. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial


Data Trial 1 Trial 2
Mass of empty crucible with lid 26.687g 26.696g
Mass of Mg metal, crucible, and lid 27.105g 27.014g
Mass of MgO, crucible, and lid  27.374g 27.218g

Mass of Mg for each trial:

Trial 1: 27.105g - 26.687g = 0.418

Trial 2: 27.014g - 26.696g = 0.318


1a. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial

1b. Determine the percent yield of MgO for your experiment for each trial

1c. Determine the average percent yield of MgO for the two trials. 


1 Expert Answer

By:

Matt W. answered • 04/01/21

Tutor
4.9 (16)

Chemical Engineer and Chemistry Enthusiast

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