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 )

Question

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 ) (A) 42 g (B) 56 g (C) (14/3) g (D) (7/4) g

Tardigrade Admin · Accepted Answer

M+X2 longrightarrow M X2 3 M X2+X2 longrightarrow M3 X8 M3 X8+ Na 2 CO 3 longrightarrow N X+ CO 2+ M 3 O 4 mole of NX = (206/103) = 2 POAC for X Atom: No. of X atom in M3X8 = No. of X Atom in NX 8 [No. of mole of M3X8] = 1 [ No. of mole of NX] No. of mole of M3X8 = [(2/8)] = (1/4) mole Now POAC for M Atom 3 [ No. of mole of M3X8 = 1 × [No. of Mole of M] ∴ 3× (1/4) = No. of mole of M weight of M atom = (3/4) × 56 = 42 grams

Q. $NX$ is produced by the following step of reactions $M + X_{2} ⟶ M X_{2}; 3 M X_{2} + X_{2} ⟶ M_{3} X_{8}; M_{3} X_{8}$ $+ N_{2} C O_{3} ⟶ NX + C O_{2} + M_{3} O_{4}$
How much $M$ (metal) is consumed to produce $206 g$ of $NX$ . (Take at. wt. of $M = 56, N = 23, X = 80$ )

$42 g$

$56 g$

$\frac{14}{3} g$

$\frac{7}{4} g$

Q. NX is produced by the following step of reactions M+X2​⟶MX2​;3MX2​+X2​⟶M3​X8​;M3​X8​ +N2​CO3​⟶NX+CO2​+M3​O4​ How much M (metal) is consumed to produce 206g of NX. (Take at. wt. of M=56,N=23,X=80 )

42g

56g

314​g

47​g

Q. $NX$ is produced by the following step of reactions $M + X_{2} ⟶ M X_{2}; 3 M X_{2} + X_{2} ⟶ M_{3} X_{8}; M_{3} X_{8}$ $+ N_{2} C O_{3} ⟶ NX + C O_{2} + M_{3} O_{4}$
How much $M$ (metal) is consumed to produce $206 g$ of $NX$ . (Take at. wt. of $M = 56, N = 23, X = 80$ )

$42 g$

$56 g$

$\frac{14}{3} g$

$\frac{7}{4} g$