Question

$Cu+HN{O}_3⟶C{u}^{2+}+N{O}_2\left(g\right)+H_{2}O$

$Zn+H^{+}+N{O}_3^{-}⟶N{H}_4^{+}+Z{n}^{2+}+H_{2}O$

The above reaction takes place when $1.5$ g of brass containing $Cu$ and $Zn$ reacts with $3MHN{O}_3$ solution. The liberated $N{O}_2$(g) was found to be $1.04$ litres at ${25}^{\circ }C$ and one atm pressure. How many mL of $3M$ $HN{O}_3$ will be required for completely reacting $1$ g of brass?

• A. $9.558$
• B. $19.108$
• C. $4.274$
• D. $6.348$

Answer 1

The correct option is A $9.558$

$1$ g brass means $0.9$ g $Cu$ and $0.1$ g $Zn$.

$\therefore$ $M_{eq.}$ of $Zn=$ $M_{eq.}$ of $HN{O}_3$

$[8e^{-}+N^{5+}⟶N^{3-},Z{n}^0⟶Z{n}^{2+}+2e^{-}]$

$\frac{0.1}{\left(\frac{65}{2}\right)}\times 1000=3\times 8\times V_1$ $⟹V_1=0.128$ mL

Also, $M_{eq.}$ of $Cu=$ $M_{eq.}$ of $HN{O}_3$

$\frac{0.9}{\left(\frac{63.6}{2}\right)}\times 1000=3\times 1\times V_2$ $⟹V_2=9.43$ mL

Thus, total volume used $=V_1+V_2=0.128+9.43$ $=9.558$ mL

Hence, the correct option is $\left(A\right)$