Question

$Ag{\left(N{H}_3\right)}_2^{+}\to {Ag}^{+}+2N{H}_3$
If $5.8g$ of $Ag{\left(N{H}_3\right)}_2^{+}$ yields $1.4g$ of ammonia, how many moles of silver will be produced?

• A. $4.4$
• B. $5.8$
• C. $0.041$
• D. $0.054$
• E. $7.2$

Answer 1

Answer: (C)

$0.041$

$Ag(N{H}_3{)}_2^{+}\to A{g}^{+}+2N{H}_3$
It is given that, $1.4$ g of $N{H}_3$ are produced by 5.8 g of $Ag(N{H}_3{)}_2^{+}$,
The number of moles is obtained by dividing mass with molar mass.
The molar masses of $Ag(N{H}_3{)}_2^{+}$ and $N{H}_3$ are 142 g/mol and 17 g/mol respectively.
The number of moles $Ag(N{H}_3{)}_2^{+}=\frac{5.8}{142}=0.0408$ moles.
The number of moles of $N{H}_3=\frac{1.4}{17}=0.0823$ moles.
0.0408 moles of $Ag(N{H}_3{)}_2^{+}$ produces 0.0823 moles of $N{H}_3$.
0.0408 moles of $Ag(N{H}_3{)}_2^{+}$, will produce the 0.0408 moles of $A{g}^{+}$ ions
which can be approximated to 0.041 moles.