# 1 g of Mg is burnt in a closed vessel which contains 0.5 g of dioxygen. what is the limiting reagent? What is the amount of Mgo formed in the reaction?

Burning of Magnesium:
2Mg + O2 $\to$ 2MgO
​Mass of Mg = 1g
Molar Mass of Mg = 24g
​Moles of Mg =

​Mass of O2 = 0.5g
Molar mass of O2 = 32g
​Moles of O2

1mol of O2 reacts with 2 moles of Mg
Therefore, 0.015mol will react with 2$×$0.015mol of Mg i.e., 0.03mol of Mg.
Since we have 0.041mol of Mg, so (0.041-0.03)mol = 0.011mol of Mg will be left over and Oxygen will be limiting reagent.

Amount of MgO formed in the reaction:
1 mol of O2 gives 2 moles of MgO, so 0.015mol of O2 will give 0.03mol og MgO.
Molar mass of MgO = 40.3g
Mass of MgO formed = molar mass $×$ moles = 40.3$×$0.03 = 1.209g

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First, you need to find the balanced equation i.e.

2Mg+O2= 2MgO

Here 1g of Mg is taken, so moles of Mg= 1/24=0.041 moles

0.5 g of O2 is taken, so moles of O2=0.5/32= 0.01 moles

For 0.01 moles of O2, 2*0.01 moles of Mg are required according to Stoichiometric Coefficients.

But we have 0.041 moles, so Mg is excess reagent whereas O2 is limiting reagent.

Moles of MgO formed = Moles of O2 * 2

0.01*2=0.02 moles of MgO. Mass of MgO= 0.02*24= 0.48 g of MgO.

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