Question

Match the weight of reactants given in column I with weight of products marked (?) given in column II.

	Column I		Column II
a.	$2H_2+O_2\to 2H_{2}O$ $1.0g$ $1.0g$ $?$	p.	0.56 g
b.	$N_2+3H_2\to 2N{H}_3$ $1.0g$ $1.0g$ $?$	q.	1.333 g
c.	$CaC{O}_3\underrightarrow{\Delta }CaO+C{O}_2$ $1.0g$ $?$	r.	1.125 g
d.	$C+2H_2\to C{H}_4$ $1.0g$ $1.0g$ $?$	s.	1.214 g

• A. $a\to r,b\to s,c\to p,d\to q$
• B. $a\to q,b\to s,c\to p,d\to r$
• C. $a\to r,b\to s,c\to q,d\to p$
• D. $a\to s,b\to r,c\to p,d\to q$

Answer 1

The correct option is A $a\to r,b\to s,c\to p,d\to q$

$a\to r$ $2H_2+O_2\to 2H_{2}O$
$1.0g$ $1.0g$
$\frac{1}{2}$ mol $\frac{1}{32}$ mol
$O_2$ is the limiting reagent : 1 mol $O_2\equiv 2$ mol $H_{2}O$
$\Rightarrow \frac{1}{32}$ mol $O_2\equiv \frac{1}{16}$ mol $\equiv \frac{18}{16}$ g of $H_{2}O$

$\Rightarrow \frac{1}{32}$ mol $O_2\equiv \frac{18}{16}$ g of $H_{2}O$
$=1.125$ g
$b\to s$ $N_2+3H_2\to 2N{H}_3$
$1.0g$ $1.0g$
$\frac{1}{28}$ mol $\frac{1}{2}$ mol
$N_2$ is the limiting reagent.
$\therefore \frac{1}{28}$ mol $N_2\equiv \frac{1}{14}$ mol $N{H}_3\equiv \frac{1}{14}\times 17g=1.214g$

$c\to p$ $CaC{O}_3\underrightarrow{\Delta }CaO+C{O}_2$
$1.0g$
$\frac{1}{100}$ mol $\frac{1}{100}$ mol $\frac{1}{100}$ mol
$=\frac{1}{100}\times 56gCaO$
$=0.56g$

$d\to q$ $C+2H_2\to C{H}_4$
$1.0g$ $1.0g$
$\frac{1}{12}$mol $\frac{1}{2}$mol
C is the limiting reagent.
$\frac{1}{12}molC=\frac{1}{12}molC{H}_4$
weight of $C{H}_4=\frac{16}{12}=1.33g$