Question

Match the followings
$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \left(I\right)& Zn\left(s\right)+2HCl\left(aq\right)\to ZnC{l}_2\left(aq\right)+H_2\left(g\right)& \left(P\right)& 50\%\text{of excess reagent left}\\ & \text{Above reaction is carried out by taking}& & \\ & \text{2 moles each of}Zn\text{and}HCl& & \\ \left(II\right)& AgN{O}_3\left(aq\right)+HCl\left(aq\right)\to AgCl\left(s\right)+HN{O}_3\left(aq\right)& \left(Q\right)& \text{22.4 L of gas at STP is liberated}\\ & \text{Above reaction is carried out by taking}& & \\ & \text{170 g}AgN{O}_3\text{and 18.25 g}HCl(Ag=108)& & \\ \left(III\right)& CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)& \left(R\right)& \text{1 moles of solid (product) obtained}\\ & \text{100 g}CaC{O}_3\text{is decomposed}& & \\ \left(IV\right)& 2KCl{O}_3\left(s\right)\to 2KCl\left(s\right)+3O_2\left(g\right)& \left(S\right)& HCl\text{is the limiting reagent}\\ & \frac{2}{3}\text{moles of}KCl{O}_3\text{decomposed}& & \\ & & \left(T\right)& \text{11.2 L of gas at STP is liberated}\\ & & \left(U\right)& 75\%\text{of excess reagent left}\end{array}$

• A. $I-P,Q,S;II-P,S;III-Q,R;IV-Q$
• B. $I-P,Q;II-T,U;III-P,Q,R;IV-P$
• C. $I-R,S;II-S,T;III-Q,T;IV-Q,R$
• D. $I-P,Q,R,S;II-Q,S;III-Q,R;IV-S$

Answer 1

The correct option is A $I-P,Q,S;II-P,S;III-Q,R;IV-Q$

$\begin{array}{cccccccc}\left(I\right)& Zn\left(s\right)& +& 2HCl\left(aq\right)& \to & ZnC{l}_2\left(s\right)& +& H_2\left(g\right)\\ \text{Initial mole}& 2& & 2& & 0& & 0\\ \text{Final mole}& (2-1=1)& & 0& & 1& & 1\end{array}$
Excess reagent left $=\frac{2-1}{2}\times 100=50\%$
Volume of $H_2=22.4\text{lit}$
Solid product obtained = 1 mole
Limiting reagent is $HCl$

$\begin{array}{cccccccc}\left(II\right)& AgN{O}_3\left(aq\right)& +& HCl\left(aq\right)& \to & AgCl\left(s\right)& +& HN{O}_3\left(g\right)\\ \text{Initial mole}& \frac{170}{170}=1& & \frac{18.25}{36.5}=\frac{1}{2}& & 0& & 0\\ & 1-\frac{1}{2}=\frac{1}{2}& & 0& & \frac{1}{2}& & \frac{1}{2}\end{array}$
Excess reagent $=\frac{1-\frac{1}{2}\times 100}{1}=50\%$
Volume of gas $=11.2\text{lit}$
Solid product $=\frac{1}{2}\text{mole}$
Limiting reagent is $HCl$

$\begin{array}{cccccc}\left(III\right)& CaC{O}_3\left(s\right)& \to & CaO\left(s\right)& +& C{O}_2\left(g\right)\\ \text{Initial mole}& \frac{100}{100}=1& & 0& & 0\\ & 0& & 1& & 1\end{array}$
Excess reagent not present
Volume of gas $=22.4\text{lit. at STP}$
Solid product is 1 mole

$\begin{array}{cccccc}\left(IV\right)& 2KCl{O}_3\left(s\right)& \to & 2KCl& +& 3O_2\left(g\right)\\ \text{Initial mole}& \frac{2}{3}& & 0& & 0\\ & 0& & \frac{2}{3}& & 1\end{array}$
No excess reagent left
Volume of gas $=22.4\text{lit}$
Solid product is $\frac{2}{3}$ mole.