Information- The chemical equation is Mg(s) + O2(g) → MgO(s)

for trial 1-
the mass of empty crucible with lid is- 26.680 g
the mass of Mg metal, crucible and lid is- 26.993 g
the mass of MgO, crucible and lid is- 27.194 g

for trial 2-
the mass of empty crucible with lid is- 26.682 g
the mass of Mg metal, crucible and lid is- 27.100 g
the mass of MgO, crucible and lid is- 27.194 g

Questions-

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

2. what is the percent yield of MgO for the experiment for each trial?

3. what is the average percent yield of MgO for the two trials?

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

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