In this question, we have the following reaction:
2 Mg + O2 -> 2 MgO
The molar ratios of the reactants and products are:
2 Mg = 1 O2
2 Mg = 2 MgO
1 O2 = 2 MgO
We have:
30 grams of Magnesium, the molar mass of Magnesium is 24.305g/mol, this is important to find the number of moles of Mg
30 grams of Oxygen gas, the molar mass is 32g/mol
Now let's check how many moles of Mg we have:
24.305g = 1 mol
30g = x moles
x = 1.23 moles of Mg in 30 grams
According to the molar ratio, if we have 1.23 moles, we need to have half of this value for O2, therefore 1.23/2 = 0.615 moles of O2
0.615 moles of O2 is the ideal value for the reaction to occur with 1.23 moles of Mg, but we need to check how many moles of O2 we actually have:
32g = 1 mol of O2
30g = x moles of O2
x = 0.937 moles of O2 in 30 grams, which means that we have more O2 than what we actually need, making O2 the excess reactant and Mg the limiting reactant.
We have 0.937 moles of O2, but we need 0.615 moles, let's check how much we have in excess:
0.937 - 0.615 = 0.322 moles, which in grams will be:
0.322 * 32 grams = 10.3 grams in excess
In order to find how much MgO is produced, we need to use the same number of moles of the limiting reactant, Mg, 1.23 moles, and also the molar mass of MgO, 40.3g/mol
40.3g = 1 mol
x grams = 1.23 moles
x = 49.6 grams of MgO is produced
Answers:
A. 49.6 grams of MgO
B. Magnesium, Mg
C. 10.3 grams of O2 in excess