NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- s Block Elements
NTA Abhyas 2020 · Chemistry · s Block Elements
The compound A on heating gives a colourless gas and a residue that...
The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of CO_{2} is bubbled through aqueous solution of B, C is formed which is recovered in the solid form. Solid C on gentle heating gives back A. The compound is
Options
- ACaCO_{3}
- BNa_{2}CO_{3}
- CK_{2}CO_{3}
- DCaSO_{4}·2H_{2}O
Correct answer
A. CaCO_{3}
Step-by-step solution
The reactions can be summarised as follows :
\text{A} \overset{Δ}{→} \text{Colourless gas} + \text{residue}
\text{Residue} + {H}_{2} \text{O} \to \text{B} \overset{\text{Excess CO} 2}{→} \text{C} \overset{Δ}{→} \text{A}
This is possible only when A is CaCO3 . The reactions are as follows :
\begin{cases}{CaCO}_{3} & \overset{Δ}{→} & {CO}_{2} & + & \text{CaO} \\ \left(\text{A}\right) & & \left(\text{Colourless gas}\right) & & \left(\text{Residue}\right)\end{cases}
\begin{cases}\text{CaO} & + & {H}_{2} \text{O} & \to & \text{Ca} {\left(\text{OH}\right)}_{2} & \mathop{→}_{\text{excess}^{\text{CO} 2} & \text{Ca} {\left({HCO}_{3}\right)}_{2} & \overset{Δ}{→} & {CaCO}_{3} \\ & & & & \left(\text{B}\right) & & \left(\text{C}\right) & & \left(\text{A}\right)\end{cases}
\text{A} \overset{Δ}{→} \text{Colourless gas} + \text{residue}
\text{Residue} + {H}_{2} \text{O} \to \text{B} \overset{\text{Excess CO} 2}{→} \text{C} \overset{Δ}{→} \text{A}
This is possible only when A is CaCO3 . The reactions are as follows :
\begin{cases}{CaCO}_{3} & \overset{Δ}{→} & {CO}_{2} & + & \text{CaO} \\ \left(\text{A}\right) & & \left(\text{Colourless gas}\right) & & \left(\text{Residue}\right)\end{cases}
\begin{cases}\text{CaO} & + & {H}_{2} \text{O} & \to & \text{Ca} {\left(\text{OH}\right)}_{2} & \mathop{→}_{\text{excess}^{\text{CO} 2} & \text{Ca} {\left({HCO}_{3}\right)}_{2} & \overset{Δ}{→} & {CaCO}_{3} \\ & & & & \left(\text{B}\right) & & \left(\text{C}\right) & & \left(\text{A}\right)\end{cases}