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How much MgO is obtained on heating 5 gm of ($MgC{O_3}$) Magnesium carbonate?a.) 2.4 gmb.) 2.38 gmc.) 2.8 gmd.) 3.28 gm

Last updated date: 20th Jun 2024
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Hint: In this reaction, the carbon dioxide is evolved and its mass is lost while the rest remains the same. Further, we know that 1 mole of a reactant forms 1 mole of product. And 1 mole of a quantity is equal to the molecular weight of a molecule. Thus, by putting the values one can easily calculate the mass obtained.

For this, first, we should know the reaction that is occurring here. What happens when Magnesium carbonate is heated that it gets converted into MgO. So, let us see the reaction.
$MgC{O_3} \to MgO + C{O_2}$

So, due to the heating of Magnesium carbonate; the carbon dioxide gas is getting evolved.
Given :
Weight of Magnesium carbonate = 5 gm
To find :
Mass of MgO formed

So, we know that Molecular mass of Magnesium carbonate = 84 g
And the molecular mass of Magnesium oxide = 40 g
Thus, if we heat 84 g of magnesium carbonate, we would get 40 g of magnesium oxide.
84 g of Magnesium carbonate = 40 g of Magnesium oxide
1 g of Magnesium carbonate = $\dfrac{{40}}{{84}}$ g of Magnesium oxide
1 g of Magnesium carbonate = 0.476 g of Magnesium oxide
5 g of Magnesium carbonate = 0.476 $\times$ 5 g of Magnesium oxide
5 g of Magnesium carbonate = 2.38 g of Magnesium oxide
So, the correct answer is “Option B”.

Note: It must be noted that the molecular mass of a substance is equal to the mass of an atom of the element multiplied by the number of atoms of that element present in the molecule. Further, 1 mole of a substance is equal to the molecular mass of that substance and also equal to Avogadro number of particles or atoms.