Question

$1\mathrm{mole}$ of Rhombic Sulphur is treated with $\mathrm{conc}.{\mathrm{HNO}}_3$. Find the mass of ${\mathrm{H}}_2\mathrm{O}$ formed.

Answer 1

Step 1: Write the balanced chemical equation between Rhombic Sulphur $\left(S_{\mathbf{8}}\right)$ and $\mathbf{conc}\mathbf{.}\mathbf{}{\mathbf{HNO}}_{\mathbf{3}}$:

${\mathrm{S}}_8+48{\mathrm{HNO}}_3\to 8{\mathrm{H}}_2{\mathrm{SO}}_4+48{\mathrm{NO}}_2+16{\mathrm{H}}_2\mathrm{O}$

From the above balanced equation, the number of moles of ${\mathrm{H}}_2\mathrm{O}$ is ${\mathrm{n}}_{\left({\mathrm{H}}_2\mathrm{O}\right)}=16\mathrm{moles}$

Step 2: Find the mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$:

$$\begin{gathered} \mathrm{Mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_2\mathrm{O}\times \mathrm{Molecular}\mathrm{weight}\mathrm{of}{\mathrm{H}}_2\mathrm{O} \\ \Rightarrow \mathrm{Mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=16\mathrm{mol}\times 18\mathrm{g}{\mathrm{mol}}^{-1} \\ \Rightarrow \mathrm{Mass}\mathrm{of}{\mathrm{H}}_2\mathrm{O}=288\mathrm{g} \end{gathered}$$

Therefore, the mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ formed when Rhombic Sulphur and $\mathbf{conc}\mathbf{.}\mathbf{}{\mathbf{HNO}}_{\mathbf{3}}$ are reacted with each other is ${\mathbf{w}}_{\left(H_{2}O\right)}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{288}\mathbf{}\mathbf{g}$.