Question

A sample of pure $CuO$ was reduced with $H_2$ gas and $H_{2}O$ formed was collected in a $44.8$ L flask containing dry $N_2$. At ${27}^{o}C$, the total pressure containing $N_2$ and $H_{2}O$ was $1.0$ atm. The relative humidity in the flask was $80\%$. The vapour pressure of water at ${27}^{o}C$ is $25$ mm. If $x$ grams of $CuO$ was reduced, then value of $10x$ is :

• A. $38$
• B. $35$
• C. $20$
• D. none of the above

Answer 1

Answer: (A)

$38$

The reaction is as follows:

$CuO+H_2\to Cu+H_{2}O\left(g\right)$
Volume of dry $N_2=44.8$ L at ${27}^{o}C$
Total pressure $(N_2+H_{2}O)=1.0$ atm
Vapour pressure of $H_{2}O=25$ mm
Humidity $=80\%$
Pressure due to $H_{2}O$ vapour$=0.8\times \frac{25}{760}$ mm

Moles of $H_{2}O$ vapour$=\frac{PV}{RT}=\frac{0.8\times \frac{25}{760}\times 44.8}{0.082\times 300}$ $=0.0478$ mol

According to the above equation,
$1$ mol of $H_{2}O$ is obtained from $1$ mol of $CuO$.
Therefore, moles of $CuO$ reduced$=0.0478$ mol.
Weight of $CuO=0.0478\times (63.5+16)=3.8$ g.