Answer:
\large \boxed{\text{150 g TiCl}_{4}}
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 189.68 79.87
TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl
m/g: 50.0
To solve this stoichiometry problem, you must
1. Theoretical yield of TiO₂
[tex]\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}[/tex]
2. Moles of TiO₂
[tex]\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}[/tex]
3, Moles of TiCl₄
The molar ratio is 1 mol TiO₂:1 mol TiCl₄.
[tex]\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}[/tex]
4. Mass of TiCl₄
[tex]\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}[/tex]
