Answer
Verified
35.1k+ 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
To get a maximum current in an external resistance class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
Other Pages
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
The strongest oxidising agent among the following is class 11 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main