Match the following : The correct match is array llll A & B & C & D...

Match the following : Match the following : The correct match is array llll A & B & C & D array

The correct match is

\begin{array}{llll}\mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D}\end{array}

Options

A $\begin{array}{llll}\text{ iv } & \text{ i } & \text{ ii } & \text{ iii }\end{array}$
B $\begin{array}{llll}\mathrm{v} & \text{ i } & \text{ ii } & \text{ iii }\end{array}$
C $\begin{array}{llll}\text{ iv } & \text{ i } & \text{ iii } & \text{ ii }\end{array}$
D $\begin{array}{llll}\text{ i } & \text{ iv } & \text{ ii }\end{array}$

A. $\begin{array}{llll}\text{ iv } & \text{ i } & \text{ ii } & \text{ iii }\end{array}$

(A)

\begin{aligned} & \underset{100 \mathrm{~g}}{\mathrm{CaCO}_3} \underset{\text{ decomposition }}{\longrightarrow} \mathrm{CaO}+\underset{22.4 \mathrm{~L}}{\mathrm{CO}_2} \\ & \because 100 \mathrm{~g} \mathrm{CaCO}_3 \text{ on decomposition gives } \\ & =22.4 \mathrm{~L} \mathrm{CO}_2 \\ & =\frac{22.4 \times 10}{100} \mathrm{LCO}_2 \\ & =2.24 \mathrm{~L} \mathrm{CO}_2 \\ & \end{aligned}

\therefore 10 \mathrm{~g} \mathrm{CaCO}_3

on decomposition will give

\begin{aligned} & =\frac{22.4 \times 10}{100} \mathrm{LCO}_2 \\ & =2.24 \mathrm{LCO}_2 \end{aligned}

(B)

\begin{aligned} & \mathrm{Na}_2 \mathrm{CO}_3 \stackrel{\text{ Excess } \mathrm{HCl}}{\longrightarrow} 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \\ & 106 \mathrm{~g} \\ & 22.4 \mathrm{~L} \\ & 106 \mathrm{~g} \mathrm{Na}_2 \mathrm{CO}_3 \text{ gives }=22.4 \mathrm{~L} \mathrm{CO}_2 \\ & 1.06 \mathrm{~g} \mathrm{Na}_2 \mathrm{CO}_3 \text{ will give } \\ & =\frac{22.4 \times 1.06}{106} \mathrm{~L} \mathrm{CO}_2 \\ & =0.224 \mathrm{~L} \mathrm{CO}_2 \\ & \end{aligned}

$\begin{aligned} & =\frac{22.4 \times 1.06}{106} \mathrm{LCO}_2 \\ & =0.224 \mathrm{~L} \mathrm{CO}_2 \end{aligned}

(C)

12 \mathrm{~g}

carbon on combustion gives

=22.4 \mathrm{~L} \mathrm{CO}_22.4 \mathrm{~g}

carbon on combustion will give

\begin{aligned} & =\frac{22.4 \times 2.4}{12} \mathrm{LCO}_2 \\ & =2 \times 2.24 \mathrm{LCO}_2 \\ & =4.48 \mathrm{LCO}_2 \end{aligned}

(D)

56 \mathrm{~g}

carbon monoxide on combustion gives

=2 \times 22.4 \mathrm{~L} \mathrm{CO}_20.56 \mathrm{~g}

carbon monoxide on combustion will give

\begin{aligned} & =\frac{2 \times 22.4 \times 0.56}{56} \mathrm{LCO}_2 \\ & =0.448 \mathrm{LCO}_2 \end{aligned}

$ Hence, A-(iv), B-(i), C-(ii), D-(iii)