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

Question

Mass of empty crucible with lid	26.689 g	26.687 g
Length of ribbon cut	4 inches	2 inches
Mass of Mg metal, crucible, and lid	27.105 g	26.990 g
Mass of MgO, crucible, and lid	27.367 g	27.182 g

Trial 1:

Trial 2:

Answer 1

Begin with a correctly balanced equation:

2Mg(s) + O2(g) ==> 2MgO(s)

Since Mg is said to limiting, the moles of Mg will determine the theoretical yield of MgO. Find moles Mg used:

mass Mg trial 1 = 27.105 g - 26.689 g = 0.416 g

mass Mg trial 2 = 26.990 - 26.687 g = 0.303 g

Calculate moles Mg used:

moles Mg trail 1 = 0.416 g Mg x 1 mol Mg / 24.31 g = 0.01712 moles Mg

moles Mg trial 2 = 0.303 g Mg x 1 mol Mg / 24.31 g = 0.01246 moles Mg

Calculate theoretical yield of MgO:

trial 1: 0.01712 mols Mg x 2 mol MgO / 2 mol Mg x 40.30 g MgO / mol = 0.690 g MgO (3 sig.figs.)

trial 2: 0.01246 mols Mg x 2 mol MgO / 2 mol Mg x 40.30 g MgO / mol = 0.502 g MgO (3 sig.figs.)