Question

How many grams of titanium (IV) oxide can be produced when 60 g of $TiC{l}_4$ is made to react with 40 g of $O_2$ as shown in the following equation? (Molar mass of titanium = 48 g/mol)
$TiC{l}_4\left(s\right)+O_2\left(g\right)\to Ti{O}_2\left(s\right)+2C{l}_2\left(g\right)$

• A. 44.8 g
• B. 75.3 g
• C. 34.5 g
• D. 25.3 g

Answer 1

The correct option is D 25.3 g

The balanced equation for the reaction of $TiC{l}_4$ with $O_2$ is as follows.
$TiC{l}_4\left(s\right)+O_2\left(g\right)\to Ti{O}_2\left(s\right)+2C{l}_2\left(g\right)$
Moles of $O_2=\frac{\text{given mass}}{\text{molar mass}}=\frac{40}{32}=1.25mol$
Moles of $TiC{l}_4=\frac{\text{given mass}}{\text{molar mass}}=\frac{60}{190}$ = 0.316 mol
1 mol of $O_2$ reacts with 1 mol of $TiC{l}_4$
1.25 mol of $O_2$ will react with 1.25 mol of $TiC{l}_4.$
Therefore, $TiC{l}_4$ is the limiting reagent.
1 mol of $TiC{l}_4$ produces 1 mol of $Ti{O}_2$
0.316 mol of $TiC{l}_4$ will produce 0.316 mol of $Ti{O}_2$
Mass of $Ti{O}_2$ formed = 0.316 mol $\times$ molar mass of $Ti{O}_2=0.316\times 80$ g
Mass of $Ti{O}_2$ formed = 25.26 g