Question

2 g magnesium is burnt in closed vessel containing 3 g oxgen calculate mole of reactant lefft over
how many grams of mgo are produced

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Answer 1

Explanation:

ANSWER

24g

Mg

+

16g

2

1

O

2

⟶MgO

1gm Mg requires

24

16

g oxygen

So, 3gm Mg requires 3×

3

2

=2gm of oxygen

So, 3gm Mg and 2gm of oxygen reacted and 1gm of oxygen left.

Answer 2

Explanation:

Balanced equation:

2Mg(s) + O2(g) → 2MgO(s)

2mol Mg react with 1 mol O2

Molar mass Mg = 24.3g/mol

Molar mass O = 16g/mol.

Molar mass O2 = 32g/mol

48.6g Mg react with 32g O2

Now you know that mass O2 is less than the mass of Mg

In the problem , you have excess O2 = Mg is limiting

But let us calculate actual mass of O2 reacted

2g Mg react with ; 2g Mg * 32gO2/ 48.6g Mg = 1.32g O2

Once again showing that Mg is limiting.