
When \[{\text{262 g}}\] of xenon (atomic mass = 131) reacted completely with \[{\text{152 g}}\] of fluorine (atomic mass = 19), a mixture of \[{\text{Xe}}{{\text{F}}_{\text{2}}}\] and \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] was produced. The molar ratio \[{\text{Xe}}{{\text{F}}_{\text{2}}}\] : \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] is :
A 1 : 2
B 1 : 4
C 1 : 1
D 1 : 3
Answer
570.9k+ views
Hint: Apply POAC. According to POAC, the number of moles (or mass) of a particular element in the reactants is equal to its number of moles (or mass) in products.
Complete answer:
Xenon is a noble gas. It is chemically unreactive as it has completed its octet. However, it does form some compounds especially with fluorine. Thus, xenon reacts with fluorine to form compounds such as xenon difluoride and xenon hexafluoride.
\[{\text{Xe + n }}{{\text{F}}_{\text{2}}} \to {\text{Xe}}{{\text{F}}_{\text{2}}} + {\text{Xe}}{{\text{F}}_{\text{6}}}\]
Calculate the number of moles for \[{\text{262 g}}\] of xenon (atomic mass = 131)
\[\dfrac{{262}}{{131}} = 2\]
The molecular weight of fluorine is 38.
\[2 \times 19 = 38\]
Calculate the number of moles of fluorine present in \[{\text{152 g}}\]
\[\dfrac{{152}}{{38}} = 4\]
Let the number of moles of \[{\text{Xe}}{{\text{F}}_{\text{2}}}\] and \[{\text{Xe}}{{\text{F}}_{\text{6}}}\]formed be ‘a’ and ‘b’ respectively.
Apply POAC
\[a \times 1 + b \times 1 = 2\] … …(1)
$\Rightarrow 2a + 6b = 8$ … …(2)
Rearrange equation (1)
$\Rightarrow b = 2 - a $… …(3)
Substitute equation (3) in equation (2)
$2a + 6b = 8$
$\Rightarrow 2a + 6(2 - a) = 8$
$\Rightarrow 2a - 6a + 12 = 8$
$\Rightarrow 4a = 4$
$\Rightarrow a = 1$ … …(4)
Substitute equation (4) in equation (3) and calculate b,
$b = 2 - a$
$\Rightarrow b = 2 - 1$
$\Rightarrow b = 1$
The molar ratio \[{\text{Xe}}{{\text{F}}_{\text{2}}}\] : \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] is 1:1.
So the option (C) is the correct option.
Note: POAC is the principle of atom conservation. This principle will help you in solving chemical equations. IF you apply this principle, then you do not need to balance the chemical equations. POAC is based on the conservation of mass.
Complete answer:
Xenon is a noble gas. It is chemically unreactive as it has completed its octet. However, it does form some compounds especially with fluorine. Thus, xenon reacts with fluorine to form compounds such as xenon difluoride and xenon hexafluoride.
\[{\text{Xe + n }}{{\text{F}}_{\text{2}}} \to {\text{Xe}}{{\text{F}}_{\text{2}}} + {\text{Xe}}{{\text{F}}_{\text{6}}}\]
Calculate the number of moles for \[{\text{262 g}}\] of xenon (atomic mass = 131)
\[\dfrac{{262}}{{131}} = 2\]
The molecular weight of fluorine is 38.
\[2 \times 19 = 38\]
Calculate the number of moles of fluorine present in \[{\text{152 g}}\]
\[\dfrac{{152}}{{38}} = 4\]
Let the number of moles of \[{\text{Xe}}{{\text{F}}_{\text{2}}}\] and \[{\text{Xe}}{{\text{F}}_{\text{6}}}\]formed be ‘a’ and ‘b’ respectively.
Apply POAC
\[a \times 1 + b \times 1 = 2\] … …(1)
$\Rightarrow 2a + 6b = 8$ … …(2)
Rearrange equation (1)
$\Rightarrow b = 2 - a $… …(3)
Substitute equation (3) in equation (2)
$2a + 6b = 8$
$\Rightarrow 2a + 6(2 - a) = 8$
$\Rightarrow 2a - 6a + 12 = 8$
$\Rightarrow 4a = 4$
$\Rightarrow a = 1$ … …(4)
Substitute equation (4) in equation (3) and calculate b,
$b = 2 - a$
$\Rightarrow b = 2 - 1$
$\Rightarrow b = 1$
The molar ratio \[{\text{Xe}}{{\text{F}}_{\text{2}}}\] : \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] is 1:1.
So the option (C) is the correct option.
Note: POAC is the principle of atom conservation. This principle will help you in solving chemical equations. IF you apply this principle, then you do not need to balance the chemical equations. POAC is based on the conservation of mass.
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