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,

Weight of Mg = 480 gms

Molecular weight of Magnesium (Mg) = 24 u and Oxigen (O)=16 u

Magnesium (Mg) reacts with oxygen to form magnesium oxide (MgO)

The reaction is

$2\text{Mg}+\text{O}_{2}=2\text{MgO}\\\\\therefore2\times24\text{ gm}+2\times32\text{ gm}=2\times(24+16)\text{ gm}\\\\\therefore48\text{ gm}+32\text{ gm}=80\text{ gm}$

As per the equation 48 gm Magnesium(Mg) and 32 gm Oxygen(O) produced 80 gm of MgO.

In other words, 48 gm magnesium required 32 gm of Oxygen for combustion.

therefore, to combustion of 480 gm magnesium required $=\frac{32}{48}\times 480\text{ gm}=320\text{ gm}$ of oxygen.

Mass of MgO formed in this reaction =$\frac{80}{48}\times480=800\text{ gm}$