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  3. Chemistry
  4. N X is produced by the following step of reactions M+X2 longrightarrow M X2 ; 3 M X2+X2 longrightarrow M3 X8 ; M3 X8 + N 2 CO 3 longrightarrow NX + CO 2+ M 3 O 4 How much M (metal) is consumed to produce 206 g of N X. (Take at. wt. of M=56, N=23, X=80 )

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Q. $N X$ is produced by the following step of reactions $M+X_2 \longrightarrow M X_2 ; 3 M X_2+X_2 \longrightarrow M_3 X_8 ; M_3 X_8$ $+ N _2 CO _3 \longrightarrow NX + CO _2+ M _3 O _4$
How much $M$ (metal) is consumed to produce $206 g$ of $N X$. (Take at. wt. of $M=56, N=23, X=80$ )

Some Basic Concepts of Chemistry

Solution:

$M+X_2 \longrightarrow M X_2$
$3 M X_2+X_2 \longrightarrow M_3 X_8$
$M_3 X_8+ Na _2 CO _3 \longrightarrow N X+ CO _2+ M _3 O _4$
mole of $NX = \frac{206}{103} = 2$
$POAC$ for $X$ Atom :
No. of $X$ atom in $M_3X_8$ = No. of $X$ Atom in $NX$
$8$ [No. of mole of $M_3X_8] = 1 $[ No. of mole of $NX$]
No. of mole of $M_3X_8 = [\frac{2}{8}] = \frac{1}{4}$ mole
Now $POAC$ for $M$ Atom
$3$ [ No. of mole of $M_3X_8 = 1 \times $[No. of Mole of $M$]
$\therefore 3\times \frac{1}{4} = $ No. of mole of $M$
weight of $M$ atom $ = \frac{3}{4} \times 56 = 42$ grams