Question

Titanium(III) chloride, a substance which speeds up the reaction forming polyethylene (a form of plastic), is made by high-temperature reaction of TiCl4 vapor with H2:
2Ti〖Cl〗_(4 ) (g)+ H_2 (g)↔2 Ti〖Cl〗_3 (s)+2HCl(g)
How many grams of TiCl4 are needed for complete reaction with 155L of H2 at 435°C and 795 mm Hg pressure?

Answer 1

Answer:

1.06 × 10³ g

Explanation:

Let's consider the following reaction

2 TiCl₄(g) + H₂(g) ↔ 2 TiCl₃(s) + 2 HCl(g)

First, we need to calculate the moles of H₂. For that, we need to convert the pressure to atm and the temperature to Kelvin.

$795mmHg \times \frac{1atm}{760mmHg} = 1.05atm$

$K = \° C + 273.15 = 435 + 273.15 = 708 K$

Then, we can calculate the moles of H₂ using the ideal gas equation.

$P \times V = n \times R \times T\\n = \frac{1.05atm \times 155L}{(0.0821atm.L/mol.K) \times 708K} = 2.80 mol$

The molar ratio of TiCl₄ to H₂ is 2:1. We can use this relation to calculate the moles of TiCl₄.

$2.80molH_2 \times \frac{2molTiCl_4}{1molH_2} = 5.60molTiCl_4$

The molar mass of TiCl₄ is 189.68 g/mol. We will use this data to calculate the mass corresponding to 5.60 moles.

$5.60mol \times \frac{189.68g}{mol} = 1.06 \times 10^{3} g$