Question

In the chemical reaction,
$K_{2}C{r}_2{O}_7+a{H}_{2}S{O}_4+bS{O}_2\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+c{H}_{2}O$. $a,b$ and $c$ are:

• A. $1,3,1$
• B. $4,1,4$
• C. $3,2,3$
• D. $2,1,2$

Answer 1

The correct option is A $1,3,1$

$K_{2}C{r}_2{O}_7+a{H}_{2}S{O}_4+bS{O}_2⟶K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+C{H}_{2}O$

	Reactant's side	Product side
$S⟶$	$a+b$	$4$
$H⟶$	$2a$	$2c$
$O⟶$	$7+4a+2b$	$4+12+c$

$\therefore a+b=4;a=c,$

$b=4-a$

$7+4a+2b=16+c$

$\Rightarrow 4a+2b=9+c$

$\Rightarrow 4a+2b=9+a$ $[\because a=c]$

$\Rightarrow 3a+2b=9$

$\Rightarrow 3a+2(4-a)=9$

$\Rightarrow 3a+8-2a=9$

$\Rightarrow a=9-8=1$

$\therefore a=1$

$b=4-a=4-1=3$

$c=a=1$

$\therefore a=1,b=3,c=1$