Question

The number of moles of ${K_2}C{r_2}{O_7}$ reduced by one mole of $S{n^{2 + }}$ ions is:
A.$1/3$
B.$1/6$
C.$2/3$
D.$3/4$

Answer 1

Hint: ${K_2}C{r_2}{O_7}$ is the formula of potassium dichromate which is an inorganic chemical reagent and mostly used as an oxidizing agent in various industrial and laboratory applications. It can be used as a sensitizer and also as an allergen.

Complete step by step answer:
${K_2}C{r_2}{O_7}$ is an ionic compound made up of $2{K^ + }$ (Potassium ions) and one polyatomic ion known as dichromate ${(C{r_2}{O_7})^{2 - }}$. Its melting point is $398^\circ C$ and its boiling point is $500^\circ C$. It is insoluble in acetone and alcohol. The reaction involves:
In ionic form –
$C{r_2}O_7^{2 - } + S{n^{ + 2}} + {H^ + } \to S{n^{ + 4}} + C{r^{ + 3}} + {H_2}O$
It includes 4 steps to balance the equation.
Step 1: Using the oxidation number, we get to know its reduction or oxidation reaction.
Reduction half- reaction: $C{r_2}O_7^{2 - } \to 2C{r^{3 + }}$ (gain of 6 electrons). [+6 converted into +3]
Oxidation half- reaction: $S{n^{2 + }} \to S{n^{4 + }}$ (loss of 2 electrons). [+2 converted into +4]
Step 2: Balance the number of electrons.
Oxidation half reaction:
$3S{n^{2 + }} \to 3S{n^{4 + }}$
[+2 converted into +4]
Step 3: Balance oxygen atoms by adding ${H_2}O$ and hydrogen by ${H^ + }$.
$C{r_2}O_7^{2 - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O$
[+6 converted into +3 ]
Step 4: Balanced reaction:
$C{r_2}O_7^{2 - } + 14{H^ + } + 3S{n^{2 + }} \to 2C{r^{3 + }} + 3S{n^{4 + }} + 7{H_2}O$.
Thus, 3 moles of $S{n^{2 + }}$ will be used to reduce 1 mole of ${K_2}C{r_2}{O_7}$.
Therefore, I mole of $S{n^{2 + }}$ reduce $1/3$ moles of ${K_2}C{r_2}{O_7}$.

So, the correct answer is OptionA .

Note:
Balanced equations are those equations in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products. ${K_2}C{r_2}{O_7}$ is a hexavalent chromium compound which is very harmful to the human body.