Question

3TiO₂ +4C+ 6Cl₂ - 3Ticl4+2CO₂+ 2CO
A vessel contains 4.32gTiO_2, 5.76g C and 7.1g Cl_2, suppose the reaction goes to completion as written, how many gram of TiCl₄ can be produced?

Answer 1

Same question try this way- only values changed

3TiO₂4C 6Cl₂ ---------> 3TiCl₄ 2CO₂ 2CO

A vessel contains 4.15 gm TiO₂, 5.67 gm C and 6.78 gm Cl₂. Suppose the reaction goes to completion how many grams of TiCl₄ can be produced?

3 TiO₂ + 4 C + 6 Cl₂ ---------> 3 TiCl₄ + 2 CO₂ + 2 CO

Mass of TiO₂ = 4.15 g

Mass of Cl₂ = 6.78 g

Molecular Mass of TiO₂ = 77 g/mol

Molecular Mass of Cl₂ = 71 g/mol

Now, (77 x 3) g = 231 g of TiO₂combines with (71 x 6) g = 426 g of Cl₂

So, 4.15 g of TiO₂ will combine with (4.15 x 426) / 231 = 7.65 g of Cl₂ .

Since, amount of Cl₂ required to completely react with TiO₂ is less therefore, Cl₂ is the limiting reagent and the amount of TiCl₄ will depend on amount of Cl₂ .

426 g of Cl₂ gives (189.8 x 3)g = 569.4 g of TiCl₄ .

Therefore, 4.15 g of TiO₂ will form (569.4 x 4.15) / 426 = 5.55 g of TiCl₄