NX is produced by the following step of $$reactionsM+X2$$→ M $$X2$$; $$3MX2+X2$$→$$M3X8$$; $$M3X8+$$ $$N2CO3$$ → $$NX+CO2+M3O4How$$ much M (metal) is consumed to produce $$206$$ g of NX. (Take$$ at. wt. of M $$=$$ $$56$$, N $$=$$ $$23$$, X $$=$$ $$80)

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NX is produced by the following step of $$reactionsM+X2$$→ M  $$X2$$; $$3MX2+X2$$→$$M3X8$$;  $$M3X8+$$  $$N2CO3$$  →   $$NX+CO2+M3O4How$$ much M (metal) is consumed to produce $$206$$ g of NX. (Take$$ at. wt. of M $$=$$ $$56$$, N $$=$$ $$23$$, X $$=$$ $$80)

Infinity Learn · Accepted Answer

Balance the given equations to calculate required quantityM          $$+$$           $$X2$$            → $$MX2$$ →(1)×$$3$$ ⇒$$3M$$          $$+$$ $$3$$          $$X2$$            → $$3MX2$$ $$3MX2+$$            $$X2$$            → $$M3X8$$→(2)M3X8+$$          $$N2CO3$$     → $$NX+CO2+M3O4$$→$$(3)By$$ balancing eq.$$(3)⇒$$M3X8+4$$          $$N2CO3$$     →$$8$$ $$NX+4CO2+M3O4From$$ the above equations ;$$3moles$$ of 'M'→$$8moles$$ of NX $$3$$×$$56g$$ of 'M'→$$8(23+80)g$$ of NX         ?         → $$206g$$ of NXmass of 'M' consumed $$=$$ $$2068(103)$$×$$3$$×$$56=42g$$

NX is produced by the following step of reactionsM+X2→ M X2; 3MX2+X2→M3X8; M3X8+ N2CO3 → NX+CO2+M3O4How much M (metal) is consumed to produce 206 g of NX. (Take at. wt. of M = 56, N = 23, X = 80)

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