Question

calculate the mass of magnesium oxide obtained when 2.4 gram of magnesium burns in oxygen in air (atomic mass of magnesium=24 and oxygen=16) find the answer

Answer 1

Answer:

Balanced chemical equation:

          2Mg + O2 ->  2MgO

       here, as we can see 2 moles of Mg need 2 moles of MgO

      Then, 2 moles Mg= 2 *  24 = 48g

      2 moles 2MgO = 2 *  40 = 80g

       48g need 80g

       24g need 40g

     then, 2.4g will need 40/10= 4gm

therefore, answer equals 4gm

Answer 2

The correct answer is 4 gm.

Given: Amount of Mg burns in air = 2.4 gm.

To Find: The mass of magnesium oxide obtained when 2.4 gram of magnesium burns in oxygen in air.

Solution:

The balanced equation is:

$2Mg_{(s)} +O_{2}_{(s)} -- > 2MgO$

Now, as the mass of Mg burnt is given so we will find the moles of Mg.

$Moles=\frac{Given mass}{molar mass}$

= $\frac{2.4}{24}$

= 0.10 mol Mg.

So, according to the reaction 2 moles of Mg is equal to 0.10 moles.

So, as in this case limiting reagent is Mg.

So, 2 moles of Mg gives  moles of MgO according to the reaction.

Now, we know moles of Mgo = 2 moles.

Molar mass of MgO = 24 + 16 = 40

$Moles=\frac{Given mass}{molar mass}$

Given mass = Moles * molar mass

= 0.10 * 40

= 4 gm.

Hence, the mass of MgO is 4 gm.

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