X + H_2SO_4 to Y (a colorless gas with a pung | vedclass

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

Question

(A) $1$. The reaction of a sulfite salt $(X = SO_3^{2-})$ with dilute $H_2SO_4$ produces sulfur dioxide gas $(Y = SO_2)$,which is a colorless gas with

Vedclass · Accepted Answer

$SO_3^{2-}, SO_2$

Vedclass · Answer

(A) $1$. The reaction of a sulfite salt $(X = SO_3^{2-})$ with dilute $H_2SO_4$ produces sulfur dioxide gas $(Y = SO_2)$,which is a colorless gas with a pungent,suffocating odor.$2$. Sulfur dioxide $(SO_2)$ acts as a reducing agent. When passed through an acidified potassium dichromate solution $(K_2Cr_2O_7 + H_2SO_4)$,it reduces the orange-colored dichromate ion $(Cr_2O_7^{2-})$ to green-colored chromium$(III)$ ions $(Cr^{3+})$.$3$. The chemical equation is: $3SO_2 + K_2Cr_2O_7 + H_2SO_4 	o K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.

$X + H_2SO_4 \to Y$ (a colorless gas with a pungent smell).
$Y + K_2Cr_2O_7 + H_2SO_4 \to$ green colored solution.
What are $X$ and $Y$?

Similar Questions

To an inorganic mixture,dilute $H_2SO_4$ is added in the cold; a colourless,odourless gas is evolved. The mixture contains:

During the qualitative analysis of $SO_{3}^{2-}$ using dilute $H_{2}SO_{4}$,$SO_{2}$ gas is evolved which turns $K_{2}Cr_{2}O_{7}$ solution (acidified with dilute $H_{2}SO_{4}$):

$A$ solution of a salt in dilute sulphuric acid imparts a deep blue colour with starch-iodine solution; it confirms the presence of which of the following?

The $Chromyl$ $Chloride$ test is performed to detect which of the following ions?

$A$ reagent consisting of $1$-naphthylamine and sulphanilic acid in acetic acid is used for the detection of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$X + H_2SO_4 \to Y$ (a colorless gas with a pungent smell).$Y + K_2Cr_2O_7 + H_2SO_4 \to$ green colored solution.What are $X$ and $Y$?