Question

1.62 g of $Ti$ reacts with an excess of $O_2$ to form $T{i}_{1.44}O$. Find the amount of the oxide formed.
(Molar mass of $Ti$ = 48 g/mol)

• A. 6 g
• B. 2 g
• C. 4 g
• D. 8 g

Answer 1

The correct option is B 2 g

Let the weight of $T{i}_{1.44}O$ formed be $x$ g.
Molar mass of $T{i}_{1.44}O=(1.44\times 48+16)=85.12g/mol$
Since the atoms of $Ti$ are conserved,
Applying POAC on $Ti$,
$1\times \text{moles of}Ti=1.44\times \text{moles of}T{i}_{1.44}O$

$\Rightarrow \frac{\text{weight of}Ti}{\text{molar mass}}=1.44\times \frac{\text{weight of}T{i}_{1.44}O}{\text{molar mass of}T{i}_{1.44}O}$

$\Rightarrow \frac{1.62}{48}=1.44\times \frac{x}{85.12}$

On solving, $x=1.99g\approx 2g$
Therefore, weight of the $T{i}_{1.44}O$ formed will be $2g$.