One third litre of x M K_2Cr_2O_7 is requir | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The balanced chemical equation for the reaction is: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ Moles of ferrous

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(B) The balanced chemical equation for the reaction is: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ Moles of ferrous ammonium sulphate $= M 	imes V = 0.1 \ M 	imes 2 \ L = 0.2 \ mol$. From the stoichiometry, $1 \ mole$ of $K_2Cr_2O_7$ reacts with $6 \ moles$ of $Fe^{2+}$. Moles of $K_2Cr_2O_7$ required $= \frac{0.2}{6} = \frac{0.1}{3} \ mol$. Molarity $x = \frac{n}{V} = \frac{0.1/3 \ mol}{1/3 \ L} = 0.1 \ M$. Thus, option $(B)$ is correct.

One third litre of $x \ M \ K_2Cr_2O_7$ is required to completely oxidise $2 \ L$ of $0.1 \ M$ ferrous ammonium sulphate in acidic medium. What is $x$?

Similar Questions

Given $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O : \Delta H = -72 \ \text{kcal}$,calculate the energy required for the formation of $1.6 \ \text{g}$ of glucose.

What is the volume (in litres) of oxygen required at $STP$ to completely convert $1.5 \ mol$ of sulphur into sulphur dioxide?

Calculate the amount of hydrogen gas required to produce $100 \text{ g}$ of ammonia by the reaction of $N_{2(g)}$ and $H_{2(g)}$ gases. (in $\text{ g}$)

For preparing $0.05 \ N \ H_2SO_4$ solution in one litre,we need $H_2SO_4$. (in $g$)

The normality of $1 \ M$ solution of $H_3PO_4$ will be.....$N$

Vedclass Test Series

Exam Paper Generator

Online Exam Module