Question

1g of Mg is burnt in a closed vessel containing 0.5g of oxygen. What is the amount of mgo formed in the reaction

Answer 1

HEYA.......


2 Mg + O2 → 2 MgO 

(1 g Mg) / (24.30506 g Mg/mol) = 0.0411 mol Mg 
(0.5 g O2) / (31.99886 g O2/mol) = 0.0156 mol O2 

0.0156 mole of O2 would react completely with 0.0156 x (2/1) = 0.0312 mole of Mg, but there is more Mg present than that, so Mg is in excess. 

((0.0411 mol Mg initially) - (0.0312 mol Mg reacted)) x (24.30506 g Mg/mol) = 0.24 g Mg left over

TYSM.#GOZMIT

Answer 2

Hey mate!

Here's your answer!!

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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