Respuesta :

znk

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

  • Convert the actual yield to the theoretical yield  
  • Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂
  • Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄
  • Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄

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]

Otras preguntas