As per the following equation, 0.217 g of Hg | vedclass

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

Question

(B) The balanced chemical equation is: $HgO + 4 I^{-} + H_2O \rightarrow HgI_4^{2-} + 2 OH^{-}$.From the stoichiometry, $1 \ mol$ of $HgO$ produces $2

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The balanced chemical equation is: $HgO + 4 I^{-} + H_2O ightarrow HgI_4^{2-} + 2 OH^{-}$.From the stoichiometry, $1 \ mol$ of $HgO$ produces $2 \ mol$ of $OH^{-}$.Number of moles of $HgO = \frac{0.217 \ g}{217 \ g \ mol^{-1}} = 0.001 \ mol$.Therefore, moles of $OH^{-}$ produced $= 2 	imes 0.001 = 0.002 \ mol$.For neutralization, $n_{H^{+}} = n_{OH^{-}}$, so $n_{HCl} = 0.002 \ mol$.Using the formula $n = M 	imes V(L)$, we have $0.002 = 0.01 	imes V(L)$.$V(L) = \frac{0.002}{0.01} = 0.2 \ L = 200 \ mL$.

Similar Questions

$0.43 \ g$ of a metal of valence $2$ was dissolved in $50 \ mL$ of $0.5 \ M \ H_2SO_4$ solution. The unreacted acid required $14.2 \ mL$ of $1 \ M \ NaOH$ solution for neutralization. The atomic weight of the metal is: (in $u$)

The molar solution of sulphuric acid is equal to:

For a gaseous reaction $Cl_{2(g)} + PCl_{3(g)} \to PCl_{5(g)}$. If $10 \ mL$ of $Cl_2$ completely react with $PCl_3$,then find out the produced volume of $PCl_5$ in $mL$.

When $CO_2\,(g)$ is passed over red hot coke,it partially gets reduced to $CO\,(g).$ Upon passing $0.5\,L$ of $CO_2\,(g)$ over red hot coke,the total volume of the gases increases to $700\,mL.$ The composition of the gaseous mixture at $STP$ is

How many $g$ of bromine will react with $21\, g$ of $C_3H_6$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module