Answer
Verified
39.3k+ views
Hint: In order to determine the amount of metal M consumed to produce 206gm of NX, the mole ratio of metal and NX shall be calculated. Moles of metal M and moles of NX can be calculated by taking the ratio of given mass and molar mass. The molar mass of NX can be calculated by adding the Atomic weight of N and X.
Complete step by step solution:
\[\begin{align}
& M+{{X}_{2}}\to M{{X}_{2}} \\
& 3M{{X}_{2}}+{{X}_{2}}\to {{M}_{3}}{{X}_{8}} \\
& {{M}_{3}}{{X}_{8}}+{{N}_{2}}C{{O}_{3}}\to NX+C{{O}_{2}}+{{M}_{3}}{{O}_{4}} \\
\end{align}\]
If equation 1 is multiplied by 3 and all equations are added, we get
\[\begin{align}
& 3M+3{{X}_{2}}\to 3M{{X}_{2}} \\
& 3M{{X}_{2}}+{{X}_{2}}\to {{M}_{3}}{{X}_{8}} \\
& {{M}_{3}}{{X}_{8}}+{{N}_{2}}C{{O}_{3}}\to NX+C{{O}_{2}}+{{M}_{3}}{{O}_{4}} \\
\end{align}\]
Resultant equation will be: \[3M+4{{X}_{2}}+{{N}_{2}}C{{O}_{3}}\to NX+C{{O}_{2}}+{{M}_{3}}{{O}_{4}}\]
Mass of NX is 206gm as given in data.
Molar Mass of NX can be calculated as given in data: $23+80=103gm$
Moles of NX can be calculated by taking the ratio of the mass of NX and molar mass of NX.
So, the number of moles of NX will be: $\dfrac{206}{103}=2 moles$
A mole ratio of Metal M and NX is 3:1
So, the number of moles of Metal M can be calculated as:
\[2moles\text{ NX}\times \dfrac{3\text{ }mole\text{ M}}{1\text{ mole NX}}\]=$6 moles M$
As 1 mole of NX is produced by 3 moles of metal M
So, 2 moles of NX will be produced by 6 moles of metal M.
As given in data, the atomic weight of metal M is $56gm$
So, 1 mole of metal M is equal to $56gm$, so $6 moles$ of metal $M$ is equal to $6moles\times 56gm=336gm$ of metal.
So (B) 336gm of metal M is consumed to produce 206gm of NX.
Note: In order to calculate mole ratio of metal M and NX, Equations should be multiplied and added. Then moles of metal M and moles of NX can be calculated by dividing given mass to molar mass. Value of 1 Mole is given in data so that it can be used to calculate the weight of 6 moles of metal M. Molar mass of NX can be calculated by adding the Atomic weight of N and X.
Complete step by step solution:
\[\begin{align}
& M+{{X}_{2}}\to M{{X}_{2}} \\
& 3M{{X}_{2}}+{{X}_{2}}\to {{M}_{3}}{{X}_{8}} \\
& {{M}_{3}}{{X}_{8}}+{{N}_{2}}C{{O}_{3}}\to NX+C{{O}_{2}}+{{M}_{3}}{{O}_{4}} \\
\end{align}\]
If equation 1 is multiplied by 3 and all equations are added, we get
\[\begin{align}
& 3M+3{{X}_{2}}\to 3M{{X}_{2}} \\
& 3M{{X}_{2}}+{{X}_{2}}\to {{M}_{3}}{{X}_{8}} \\
& {{M}_{3}}{{X}_{8}}+{{N}_{2}}C{{O}_{3}}\to NX+C{{O}_{2}}+{{M}_{3}}{{O}_{4}} \\
\end{align}\]
Resultant equation will be: \[3M+4{{X}_{2}}+{{N}_{2}}C{{O}_{3}}\to NX+C{{O}_{2}}+{{M}_{3}}{{O}_{4}}\]
Mass of NX is 206gm as given in data.
Molar Mass of NX can be calculated as given in data: $23+80=103gm$
Moles of NX can be calculated by taking the ratio of the mass of NX and molar mass of NX.
So, the number of moles of NX will be: $\dfrac{206}{103}=2 moles$
A mole ratio of Metal M and NX is 3:1
So, the number of moles of Metal M can be calculated as:
\[2moles\text{ NX}\times \dfrac{3\text{ }mole\text{ M}}{1\text{ mole NX}}\]=$6 moles M$
As 1 mole of NX is produced by 3 moles of metal M
So, 2 moles of NX will be produced by 6 moles of metal M.
As given in data, the atomic weight of metal M is $56gm$
So, 1 mole of metal M is equal to $56gm$, so $6 moles$ of metal $M$ is equal to $6moles\times 56gm=336gm$ of metal.
So (B) 336gm of metal M is consumed to produce 206gm of NX.
Note: In order to calculate mole ratio of metal M and NX, Equations should be multiplied and added. Then moles of metal M and moles of NX can be calculated by dividing given mass to molar mass. Value of 1 Mole is given in data so that it can be used to calculate the weight of 6 moles of metal M. Molar mass of NX can be calculated by adding the Atomic weight of N and X.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
The length of a potentiometer wire is 10m The distance class 12 physics JEE_MAIN
Other Pages
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
Iodoform can be prepared from all except A Acetaldehyde class 12 chemistry JEE_Main
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
448 litres of methane at NTP corresponds to A 12times class 11 chemistry JEE_Main
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main