What theoretical mass of magnesium oxide is produced when 82.56 grams of magnesium reacts with excess oxygen?

Answer 1

First write the balanced chemical equation:

2 Mg + O2 --> 2 MgO.

Now solve for the theoretical amount of MgO formed:

(82.56 g Mg / 1) * (1 mol Mg / 24.305 g Mg) * (2 mol MgO / 2 mol Mg) * (40.3044 g MgO / 1 mol MgO) = 136.9 g MgO.

Answer 2

First you need to write a balanced equation. You are given that magnesium combines with oxygen to form magnesium oxide.

Unbalanced: Mg + O2 ---> MgO

Balanced: 2Mg + O2 ---> 2MgO

Givens:

2.50 grams Mg (Molecular mass 24.3 g)

2.50 grams O2 (Molecular mass 32.0 g)

Molecular mass of MgO: 40.3 g

Mole ratio 2:1:2 (Mg:O2:MgO)

Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.2.50 g Mg × (2 moles) / (44.0 g) = 1.25 moles Mg 2.50 g O2 / (32.0 g) = .0781 moles O2 Because oxygen is the limiting reactant (less of it than magnesium), the amount of MgO produced can only be as much as the oxygen. Take the amount of oxygen and use stoichiometry to find the amount of MgO produced in grams. 2.50 g Mg / (44.0 g) × (40.3 g) = 2.29 grams H2O

Answer 3

Answer

2.70 x 2/2 x gfm of magnesium oxide = your answer is 4.

Answer 4

Four (4) moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen.

Answer 5

The reaction is;
2 Mg + O2 = 2 MgO

Answer 6

Four

Answer 7

2 moles