Question

Mole of MgO produced by the reaction of 2.4g of Mg with 1.6 g O is__?

Answer 1

Explanation:

This problem is a limiting reactant type of problem. Basically you need to solve the problem twice. But first, you must have a balanced equation for the reaction. For this reaction the equation is:

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

Now, starting with the magnesium, you must calculate the moles (and/or grams) of MgO which could be formed from that:

2.4 g Mg / 24 g/mol = 0.10 mol Mg

0.10 mol Mg ( 2 moles MgO / 2 moles Mg) = 0.10 mol MgO

Now, beginning with 0.25 mol of O2:

0.25 mol O2 X ( 2 mol MgO / 1 mol O2) = 0.50 mol MgO

The amount of magnesium that you are giving limits the amount of MgO that you could possibly form. So, this reaction can only form 0.10 mol of MgO. The mass of 0.10 mol of MgO is:

0.10 mol MgO ( 40.0 g/mol) = 4 grams MgO

Answer 2

0.1 moles of Magnesium Oxide will be produced.

Given,

Mass of Mg = 2.4 g

Mass of O₂ = 1.6 g

To Find,

Moles of MgO produced

Solution,

The reaction between Magnesium and Oxygen to produce Magnesium Oxide can be given as follows -

2Mg + O₂ → 2MgO

According to the stoichiometry -

2 moles of Magnesium and 1 mole of Oxygen produce 2 moles of Magnesium Oxide

Number of Moles = $\frac{Given mass}{Molar mass}$

Now we also know that -

The molar mass of Mg = 24 g

The molar mass of O₂ = 32 g

The moles of Mg = $\frac{2.4}{24}$

The moles of Mg = 0.1 mol

The moles of O₂ = $\frac{1.6}{32}$

The moles of O₂ = 0.05 mol

By using stoichiometry -

0.1 mole of Magnesium would require 0.05 moles of Oxygen to produce 0.1 moles of Magnesium Oxide.

Thus, both reactants are in the required condition.

Hence, 0.1 moles of Magnesium Oxide will be produced.

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