Question

How many grams of O2 is required for combustion of 480 grams of Mg? Find the mass of Mgo formed in this reaction.(Mg=24u,O=16u).

Answer 1

Given :

The required balanced chemical equation is :

2Mg     +  O2 ---->2MgO

molecular mass of Mg= 24u

molecular mass of O2= 32u

Molecular mass of MgO= 24+16=40u

48 g of Mg reacts with 32g of O2 to form 80g of MgO

How many grams of O2 is required for combustion of 480 grams of Mg

48 g of Mg------> 32g of O2

480 of Mg ---->?

= 480x32/48

=320 g of O2

so 320 g of oxygen is required for combustion of 480 grams

Sum of reactants in this reaction is 480+320 =800g

so  product MgO mass is also 800g according to the law of conservation of mass

Answer 2

given, mass of Mg = 480 grams

so, mole of Mg = 480/24 = 20

combustion of magnesium is shown by chemical reaction ..

$2Mg+O_2\rightarrow 2MgO$

in this reaction, it is clear that 1 mole of O2 is required for combustion for 2 moles of Mg.

so, 10 moles of O2 is required for combustion for 20 moles of Mg.

hence, weight of O2 is required = mole of O2 × molecular weight

= 10 × 32 = 320 grams

also we see that, 2 mole of Mg formed 2 mole of MgO

so, 20 moles of Mg formed 20 moles of MgO

hence, weight of MgO = number of mole of MgO × molecular weight

= 20 × (24 + 16)

= 20 × 40 = 800 grams