When 262 g of xenon (atomic mass = 131) reac | vedclass

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

Question

(C) The chemical reaction is: $2Xe + 4F_2 \longrightarrow XeF_2 + XeF_6$Initial moles of $Xe = \frac{262}{131} = 2 \ mol$.Initial moles of $F_2 = \fra

Vedclass · Accepted Answer

$1 : 1$

Vedclass · Answer

(C) The chemical reaction is: $2Xe + 4F_2 \longrightarrow XeF_2 + XeF_6$Initial moles of $Xe = \frac{262}{131} = 2 \ mol$.Initial moles of $F_2 = \frac{152}{38} = 4 \ mol$.Let $x$ be the moles of $XeF_2$ and $y$ be the moles of $XeF_6$ formed.According to the law of conservation of mass (atoms):For $Xe$: $x + y = 2$For $F$: $2x + 6y = 2 	imes 4 = 8$Solving the equations: $x + y = 2$ and $x + 3y = 4$.Subtracting the first from the second: $2y = 2$,so $y = 1$.Substituting $y = 1$ into $x + y = 2$,we get $x = 1$.Thus,the molar ratio $XeF_2 : XeF_6$ is $1 : 1$.

When $262 \ g$ of xenon (atomic mass $= 131$) reacted completely with $152 \ g$ of fluorine (atomic mass $= 19$),a mixture of $XeF_2$ and $XeF_6$ was produced. The molar ratio $XeF_2 : XeF_6$ is

Similar Questions

What is the value of percent atom economy if the formula weight of the product is $46 \ u$ and the sum of the formula weight of all reactants is $92 \ u$ (in $\%$)?

$0.7 \ g$ of $Na_2CO_3 \cdot xH_2O$ were dissolved in water and the volume was made to $100 \ mL$. $20 \ mL$ of this solution required $19.8 \ mL$ of $N/10 \ HCl$ for complete neutralization. The value of $x$ is:

Hydrochloric acid solutions $A$ and $B$ have concentrations of $0.5 \,N$ and $0.1 \,N$ respectively. The volumes of solutions $A$ and $B$ required to make $2 \,L$ of $0.2 \,N$ hydrochloric acid are:

The volume of gas at $NTP$ produced by $100 \ g$ of $CaC_2$ with water is $:- CaC_2 + 2 H_2O \rightarrow Ca(OH)_2 + C_2H_{2(g)}$ (in $L$)

The weight of pure $NaOH$ required to prepare $250 \ cm^3$ of $0.1 \ N$ solution is ......... $g$. (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module