The ratio of amounts of H_2S needed to precip | vedclass

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

Question

(B) For $AgNO_3$: The reaction is $2AgNO_3 + H_2S \rightarrow Ag_2S + 2HNO_3$.$2 \ mol$ of $AgNO_3$ react with $1 \ mol$ of $H_2S$.Amount of $AgNO_3 =

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

(B) For $AgNO_3$: The reaction is $2AgNO_3 + H_2S ightarrow Ag_2S + 2HNO_3$.$2 \ mol$ of $AgNO_3$ react with $1 \ mol$ of $H_2S$.Amount of $AgNO_3 = 100 \ mL 	imes 1 \ M = 100 \ mmol$.$H_2S$ required $= \frac{100}{2} = 50 \ mmol$.For $CuSO_4$: The reaction is $CuSO_4 + H_2S ightarrow CuS + H_2SO_4$.$1 \ mol$ of $CuSO_4$ reacts with $1 \ mol$ of $H_2S$.Amount of $CuSO_4 = 100 \ mL 	imes 1 \ M = 100 \ mmol$.$H_2S$ required $= 100 \ mmol$.Ratio of $H_2S$ required $= \frac{50}{100} = \frac{1}{2}$ or $1:2$.

The ratio of amounts of $H_2S$ needed to precipitate all the metal ions from $100 \ mL$ of $1 \ M$ $AgNO_3$ and $100 \ mL$ of $1 \ M$ $CuSO_4$ will be

Similar Questions

Write the reaction of aluminum with caustic soda and determine how many moles of $H_2$ gas will be produced at $STP$ from $54 \ g$ of $Al$. (in $mol$)

The volume in $mL$ of $0.1 \ M$ solution of $NaOH$ required to completely neutralize $100 \ mL$ of $0.3 \ M$ solution of $H_3PO_3$ is

If combustion of $4 \ g$ of $CH_4$ liberates $2.5 \ kcal$ of heat,the heat of combustion of $CH_4$ is ..... $kcal/mol$.

$50 \ mL$ of $0.5 \ M$ oxalic acid is needed to neutralize $25 \ mL$ of sodium hydroxide solution. The amount in grams of $NaOH$ in $50 \ mL$ of the given sodium hydroxide solution is:

An average person needs about $10000 \ kJ$ energy per day. The amount of glucose (molar mass $= 180.0 \ g \ mol^{-1}$) needed to meet this energy requirement is $..... \ g$.
(Nearest integer)
(Use : $\Delta_{c} H(\text{glucose}) = -2700 \ kJ \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module