Question

How many gram of O2 is required for combustion 480grand of mg ? find the mass of mgo in this reacton (mg=24u, O=16u

Answer 1

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.