Question

How many grams of titanium (IV) oxide can be produced when 80 g of $TiC{l}_4$ is made to react with 20 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. 88 g
• B. 130 g
• C. 34 g
• D. 50 g

Answer 1

The correct option is C 34 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 $O_2=\frac{\text{given mass}}{\text{molar mass}}=\frac{20}{32}=0.625$ moles
Moles of $TiC{l}_4=\frac{\text{given mass}}{\text{molar mass}}=\frac{80}{190}$ = 0.421 moles
1 mole of $O_2$ reacts with 1 mole of $TiC{l}_4$
0.625 moles of $O_2$ will react with 0.625 mole of $TiC{l}_4.$
Therefore, $TiC{l}_4$ is the limiting reagent.
1 mole of $TiC{l}_4$ produces 1 mole of $Ti{O}_2$
0.421 moles of $TiC{l}_4$ will produce 0.421 moles of $Ti{O}_2$
Mass of $Ti{O}_2$ formed = 0.421 moles $\times$ molar mass of $Ti{O}_2=0.421\times 80$ g
Mass of $Ti{O}_2$ formed = 33.68 g ~ 34 g