Question

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

Answer 1

Answer:

Explanation:

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 =$\frac{80}{48}\times480=800\text{ gm}$

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