$1.44 \mathrm{~g}$ of titanium (At. mass $=48$) reacted with excess of $\mathrm{O}_{2}$ and produce $\mathrm{x~g}$ of non stoichiometric compound $\mathrm{Ti}_{1.44} \mathrm{O}$. The value of $\mathrm{x}$ is:

Question

Aicharya · Accepted Answer

$\begin{array}{l}
\quad \mathrm{Ti}+\mathrm{O}_{2} & \longrightarrow & \mathrm{Ti}_{1.44} \mathrm{O} \\
\frac{1.44}{48} \text{ mole} & & \frac{\mathrm{x}}{48(1.44)+16} \text{ mole}
\end{array}$
$\therefore \frac{1.44}{48}=\frac{1.44 \mathrm{x}}{48(1.44)+16}$
$\mathrm{x}=1.77 \mathrm{~g}$

$1.44 \mathrm{~g}$ of titanium (At. mass $=48$) reacted with excess of $\mathrm{O}_{2}$ and produce…

Solution