Question

How many gram of o2 is required for combustion 480 g of mg find the mass of mgo formed in this reaction (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

Answer 2

Given: mass of Magnesium is 480 grams

To find: the mass of MgO formed in this reaction.

Solution:

• We know the formula for moles, which is

              no. of moles = given mass / molar mass

• So,

              no. of mole of Mg = 480/24

= 20

• Lets write the chemical equation involved :

                  2Mg + O2 ------> 2MgO

• In this reaction, we can see:

• 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.

• So we can find the weight of ozygen from above information:

                 mass (O2) =  molecular mass x no. of mole of (O2)

                 = 32 x 10 = 320 grams.

• Now, here 2 mole of Mg can formed 2 mole of MgO........(from above)

• So, 20 moles of Mg can formed 20 moles of MgO.

• So, mass (MgO) = number of mole (MgO) × molecular mass

                  = 20 × (24 + 16)

                  = 20 × 40 = 800 grams

Answer:

               The mass of MgO formed in this reaction is 800 grams.