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# What is the mass of magnesium oxide when you burn 4 grams of magnesium?

Updated: 8/11/2023 Wiki User

13y ago

Mg grams -> (use Mg's molar mass) -> Mg moles -> (use ratio of moles - use balanced equation) -> MgO moles -> (use MgO's molar mass) -> grams MgO

set up the equation:

Mg + O2 --> MgO

(we know the product is MgO and not MgO2 because magnesium has a charge of 2+ while oxygen has a charge or 2-)

balance the equation:

2Mg + O2 --> 2MgO

Molar mass of Mg: 24.31 g/mol

Molar mass of MgO: 44.30 g/mol (add the molar mass of Magnesium - 24.31g/mol and the molar mass of Oxygen - 15.99g/mol together)

(use Periodic Table to find these)

7.0 grams of Mg

To find the moles of Magnesium you use the molar mass of Mg.

(7.0 g Mg)*(1 mol Mg / 24.31 g Mg) =0.2879 moles Mg

notice how the grams cancel to leave you with moles - remember dividing by a fraction is the same as multiplying by the reciprocal

Now use the balanced equation's coefficients and the moles of Mg to determine the number of moles of MgO present.

2Mg + O2 --> 2MgO

2 moles Mg : 2 moles MgO -> divide both sides by 2 and it obviously becomes a 'one to one' ratio. This means that the number of moles of Mg is equal to the number of moles of MgO. This means that there are 0.2879 moles of MgO.

Now that we know MgO's molar mass and the number of moles of MgO we have, the grams of MgO produced can be determined.

(0.2879 moles MgO)*(44.30 g MgO / 1 mol MgO) = 12.75 grams MgO Wiki User

14y ago   Wiki User

11y ago

Let's start by writing a chemical equation for the reaction.

Mg (s) + O2 (g) ----> MgO (s)

According to the Law of Conservation of Mass, the amount of moles of each reagent before the reaction must equal the number of moles of each original reagent after the reaction. We must balance the equation:

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

We assume that the reaction occurs in excess oxygen and that all 8.0 grams of magnesium is completely burned. We then use stoichiometry to find the amount of MgO formed.

(8.0g Mg)(1 mol Mg / 24.3050g Mg)(2 mol MgO / 2 mol Mg)(40.3044g MgO / 1 mol MgO) = 13.3g MgO is formed

So overall 13.3g MgO is formed   Wiki User

6y ago

You might want to know why 1.99 g is the answer, or how to calculate it. This answer is arrived at as follows:2Mg + O2 ==> 2MgO ... balanced chemical equation

1.2 g Mg x 1 mole/24.3 g = 0.049 moles Mg

Since 2 moles Mg produces 2 moles MgO, this is a 1:1 mole ratio, thus...

0.049 moles Mg will produce 0.049 moles MgO

0.049 moles MgO x 40.3 g/mole = 1.98 grams = 2.0 grams (to 2 significant figures)   Wiki User

13y ago

Balanced equation.

2Mg + O2 >> 2MgO

8 grams MgO ( 1mol MgO/40.31g )(1 mol O/2 mol MgO )(16g O/1mol O) = 1.59 grams   Wiki User

13y ago

2Mg + O2 ----->2MgO

48g Mg gives 80 grams of MgO

4g Mg of gives 80/48 x 4= 80/12=20/3= 6.67g of MgO   Wiki User

6y ago

The magnesium oxide (MgO) obtained theoretically is 1,99 g.   Earn +20 pts  