Question

$1.5$ g of brass containing $Cu$ and $Zn$ reacts with $3.0M$ $HN{O}_3$ solution. The following reactions take place:
$Cu+HN{O}_3\to C{u}^{2+}+N{O}_2\left(g\right)+H_{2}O$
$Zn+H^{⨁}+N{O}_3^{⊝}\to N{H}_4^{⨁}+Z{n}^{2+}+H_{2}O$
The liberated $N{O}_2\left(g\right)$ was found to be $1.04$ L at ${25}^{o}C$ and $1$ atm.
a. Calculate the percentage composition of brass.
b. How many mL of $3.0M$ $HN{O}_3$ will be required for complete reaction with $1.0$ g of brass?

• A. a: $\%$ of $Zn=10.25\%$,$\%$ of $Cu=89.75\%$
  b: $20.11mL$
• B. a: $\%Zn=12\%$, $\%$ of $Cu=88\%$
  b: $204mL$
• C. a: $\%Zn=8\%$,$\%$ of $Cu=92\%$
  b: $20.11mL$
• D. None of these

Answer 1

The correct option is A a: $\%$ of $Zn=10.25\%$,$\%$ of $Cu=89.75\%$

b: $20.11mL$
$Cu\to C{u}^{2+}+2e^{-}$

$[\underset{\underset{x=5}{1+x-6=0}}{H^{+}+e^{-}+HN{O}_3}\to \underset{\underset{x=4}{x-4=0}}{N{O}_2+H_{2}O}]\times 2$

$Cu+2H^{⨁}+2HN{O}_3\to 2N{O}_2+2H_{2}O+C{u}^{2+}$

$Cu+4HN{O}_3\to 2N{O}_2+2H_{2}O+C{u}^{2+}+2N{O}_3^{⊝}$
or
$Cu+4HN{O}_3\to 2N{O}_2+Cu(N{O}_3{)}_2+2H_{2}O$

$1$ mol of $Cu=4$mol of $HN{O}_3=2$ mol of $N{O}_2$

mole of $N{O}_2=\frac{PV}{RT}=\frac{1\times 1.04}{0.082\times 298}=0.0425$
mole of $Cu=\frac{0.0425}{2}=\frac{0.0425}{2}\times 63.5=1.3462g$

Weight of $Zn=1.5-1.3462=0.1538g$

$\%Zn=10.25\%$

$\%$ of $Cu=89.75\%$

$[Zn\to Z{n}^{+2}+2e^{-}]\times 4$

$\underset{\underset{x=5}{x-6=-1}}{8e^{-}+N{O}_3^{⊝}+10H^{⨁}}\to \underset{\underset{x=-3}{x+4=+1}}{N{H}_4^{⨁}+3H_{2}O}$
or
$4Zn+10HN{O}_3\to N{H}_4^{⨁}+4Z{n}^{2+}+3H_{2}O$
or
$4Zn+10HN{O}_3\to N{H}_{4}N{O}_3+4Zn(N{O}_3{)}_2+3H_{2}O$

$4$ mol of $Zn=10$ mol of $HN{O}_3$

Moles of $HN{O}_3={\left(\frac{0.8975}{63.5}\times 4\right)}_{Cu}+{\left(\frac{0.1025}{65}\times \frac{10}{4}\right)}_{Zn}$
$=3\times V$

$V=0.2011L=20.11mL$