Question

2.6 g of a mixture of calcium carbonate and magnesium carbonate is strongly heated to the constant weight of 1.3 g. The atomic weights of calcium and magnesium are 40 & 24 respectively. State which of the following weights expresses the weight of calcium carbonate in the original mixture?
A. 980 mg
B. 400 mg
C. 1.75 g
D. 0.74 g

Answer 1

Hint: To find the mass of A & B in the given question, the mole concept is used. The mole concept is a simple & convenient method for expressing the amount of a substance in a chemical application. It is the best option for expressing the amount of reactants & products consumed or formed during any chemical reaction. This quantity is sometimes referred to as the chemical amount. As we know the number of moles (n) = \[\dfrac{{\rm{m}}}{{{\rm{MM}}}}\]

Complete Step by Step Solution:
Let the weight of the mixture of calcium carbonate & magnesium carbonate=
\[{{\rm{M}}_{{\rm{CaC}}{{\rm{o}}_{\rm{3}}}}}{\rm{ + }}{{\rm{M}}_{{\rm{MgC}}{{\rm{o}}_{\rm{3}}}}}{\rm{ = 2}}{\rm{.6g}}\]
Let the weight of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\]in the mixture = \[{\rm{x}}\]
The weight of \[{\rm{MgC}}{{\rm{O}}_{\rm{3}}}\]will be = \[{\rm{(2}}{\rm{.6 - x)}}\]
According to the statement \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\]on heating produces \[{\rm{CaO}}\] and \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]as follows:

Similarly, On heating \[{\rm{MgC}}{{\rm{O}}_{\rm{3}}}\]produces \[{\rm{MgO}}\]and \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]is evolved.

Given that fixed mass that is produced = 1.3g
\[{{\rm{m}}_{{\rm{CaO}}}}{\rm{ + }}{{\rm{m}}_{{\rm{MgO}}}}{\rm{ = 1}}{\rm{.3g (Residue) - - - - - - - - - - - (i)}}\]
Given that the atomic weights of calcium and magnesium are 40 & 24 respectively.
\[{{\rm{n}}_{{\rm{CaO}}}}{\rm{ = }}\dfrac{{\rm{x}}}{{{\rm{100}}}}\] \[{{\rm{n}}_{{\rm{MgO}}}}{\rm{ = }}\dfrac{{{\rm{(2}}{\rm{.6 - x)}}}}{{{\rm{84}}}}\]
Where, \[{\rm{n = }}\dfrac{{{\rm{Given Mass}}}}{{{\rm{Molar Mass}}}}\]
\[{{\rm{m}}_{{\rm{CaO}}}}{\rm{ = }}\dfrac{{\rm{x}}}{{{\rm{100}}}}{\rm{ \times 56}}\] \[{{\rm{m}}_{{\rm{MgO}}}}{\rm{ = }}\dfrac{{{\rm{(2}}{\rm{.6 - x)}}}}{{{\rm{84}}}}{\rm{ \times 40}}\]
On putting these values in equation (i)
\[\dfrac{{\rm{x}}}{{{\rm{100}}}}{\rm{ \times 56 + }}\dfrac{{{\rm{(2}}{\rm{.6 - x)}}}}{{{\rm{84}}}}{\rm{ \times 40 = 1}}{\rm{.3g}}\]
\[\begin{array}{l}{\rm{0}}{\rm{.56 + }}\dfrac{{{\rm{(2}}{\rm{.6 - x)}}}}{{{\rm{84}}}}{\rm{ \times 40 = 1}}{\rm{.3g}}\\{\rm{47}}{\rm{.04 \times 104 - 40x = 109}}{\rm{.2g }}\\{\rm{7}}{\rm{.04x = 5}}{\rm{.2}}\\{\rm{x = 0}}{\rm{.74g}}\end{array}\]

Hence, Option (D) is the correct answer. The weight of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] in original mixture is 0.74 g

Note: Try to calculate molecular mass of required compounds errorless.
(Add up the atomic masses of the elements according to the molecular formula)
Molecular mass of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}{\rm{ = 40 + 12 + 16 \times 3 = 100g}}\]
Molecular mass of \[{\rm{CaO = 40 + 16 = 56g}}\]
Molecular mass of \[{\rm{MgC}}{{\rm{O}}_{\rm{3}}}{\rm{ = 24 + 12 + 16 \times 3 = 84g}}\]
Molecular mass of \[{\rm{MgO = 24 + 16 = 40g}}\]