Question

Calculate the volume of hydrogen sulfide which will react with $10\mathrm{g}$ of lead nitrate. $\left[\mathrm{Pb}=207,\mathrm{N}=14,\mathrm{O}=16,\mathrm{molar}\mathrm{volume}\mathrm{of}\mathrm{gas}\mathrm{is}22.4{\mathrm{dm}}^3\mathrm{at}\mathrm{STP}\right]$

Answer 1

Step 1: Given information

• Mass of lead nitrate $\left(\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\right)=10\mathrm{g}$
• Hydrogen sulfide reacts with lead nitrate.
• Molar volume of Hydrogen sulfide gas at STP (Standard temperature and pressure)$=22.4{\mathrm{dm}}^3$

Step 2: Reaction involved

• Hydrogen sulfide $\left({\mathrm{H}}_2\mathrm{S}\right)$ is passed through lead nitrate $\left(\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\right)$ solution to form nitric acid $\left({\mathrm{HNO}}_3\right)$ and lead sulfide $\left(\mathrm{PbS}\right)$.
• The reaction is as follows:

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{S}\to {\mathrm{HNO}}_3+\mathrm{PbS}$

Step 3: Balanced chemical reaction

• The balanced chemical reaction is as follows:

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{S}\to 2{\mathrm{HNO}}_3+\mathrm{PbS}$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2& =& 207\mathrm{g}+2\times \left(14\mathrm{g}+16\mathrm{g}\times 3\right)\\ & =& 331\mathrm{g}\end{array}$

Step 4: Calculate the volume of Hydrogen sulfide

• $331\mathrm{g}$ of lead nitrate reacts with 1 mole or $22.4{\mathrm{dm}}^3$ of hydrogen sulfide at STP.
• Thus,$10\mathrm{g}$ of lead nitrate will react with $\frac{22.4{\mathrm{dm}}^3}{331\mathrm{g}}\times 10\mathrm{g}=0.677{\mathrm{dm}}^3$ of hydrogen sulfide.

Therefore, $10\mathrm{g}$ of lead nitrate will react with $0.677{\mathrm{dm}}^3$ of hydrogen sulfide.