Based in Magnesium being the limited reactant, calculate the theoretical yield and Percent yield. Then determine the average percent yield in the two trials.

Question

Data	Trial 1	Trial 2
Mass of empty crucible with lid	26.694g	26.693g
Mass of Mg metal, crucible, and lid	26.953g	27.002g
Mass of MgO, crucible, and lid	27.117g	27.198g

1. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for each trial. 
2. Trial 1:
3. Trial 2:

1. Determine the percent yield of MgO for your experiment for each trial.  
2. Trial 1:
3. Trial 2:
4. Determine the average percent yield of MgO for the two trials.

Answer 1

Reaction is 2 Mg + O₂ --> 2 MgO, so mole ratio Mg:MgO is 2:2.

Trial 1) 0.259g Mg * (1 mol Mg /24.305g Mg) * (2 MgO/2 Mg mole ratio) * (40.304g MgO /1 mol MgO) = 0.429₅ g MgO theoretical.

Trial 2) 0.309g Mg * (same conversions) = 0.512₄ g MgO theoretical. [4th digit is retained for precision, is not significant]

Percent yield = 100 * actual/theoretical: Trial 1) 100 * 0.423/0.429₅ = 98.4₈ %. Trial 2) 100 * 0.505/0.512₄ = 98.5₆ %.

2) Average percent yield of MgO is 98.5% with 3 sig figs.