Q3. 1 gm of Mg is burnt in a closed vessel which contains 0.5 g of oxy

1.	Q3. 1 gm of Mg is burnt in a closed vessel which contains 0.5 g of oxygen gas.(i) Which is the limiting reactant? (ii)What is the mass of MgO formed in this reaction? (iii) Calculate themass of excess reactant left unreacted
Answer» 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, 20.01 moles of Mg are required according to Stoichiometric Coefficients. But we have 0.041 moles, so Mg is excess reagent whereas O2is limiting reagent. Moles of MgO formed = Moles of O2 2 0.012=0.02 moles of MgO. Mass of MgO= 0.0224= 0.48 g of MgO.

Q3. 1 gm of Mg is burnt in a closed vessel which contains 0.5 g of oxygen gas.(i) Which is the limiting reactant? (ii)What is the mass of MgO formed in this reaction? (iii) Calculate themass of excess reactant left unreacted

Answer»

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 O2is 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.