Find the theoretical and per cent yeilds

Question

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

Data	Trial 1	Trial 2
Mass of empty crucible with lid	26.698g	26.697g
Mass of Mg metal, crucible, and lid	27.000g	26.998g
Mass of MgO, crucible, and lid	27.191g	27.191g

Answer 1

2Mg + O₂ ==> 2MgO .. balanced equation for the supposed reaction taking place

Trial 1:

mass Mg used = 27.000 g - 26.698 g = 0.302 g

moles Mg used = 0.302 g x 1 mol Mg / 24.31 g = 0.012423 moles

theoretical yield of MgO = 0.012423 mols Mg x 2 mol Mg / 2 mol Mg x 40.30 g / mol MgO = 0.501 g

actual yield of MgO = 27.191 g - 26.698 g = 0.493 g

% yield = actual / theoretical (x100%) = 0.493 g / 0.501 g (x100%) = 98.4%

Trial 2:

mass Mg used = 26.998 g - 26.697 g = 0.301 g

moles Mg used = 0.301 g x 1 mol Mg / 24.31 g = 0.01238 moles

theoretical yield of MgO = 0.01238 mols Mg x 2 mol MgO / 2 mol Mg x 40.30 g / mol MgO = 0.499 g

actual yield of Mg O = 27.191 g - 26.697 g = 0.494 g

% yield = actual / theoretical (x100%) = 0.494 g / 0.499 g (x100%) = 99.0%

Average % yield = (98.4% + 99.0%) / 2 = 98.7%