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?
2Mg + O2 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 20.015mol 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 = 40.30.03 = 1.209g
First, you need to find the balanced equation i.e.
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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