Answer:
The answer to your question is 1124.8 g of TiCl₄
Step-by-step explanation:
Data
TiCl₄ = ?
Volume of H₂ = 170 l
Temperature = 450°C
Pressure = 785 mm Hg
Process
1.- Convert temperature to °K
Temperature = 273 + 450
Temperature = 723 °K
2.- Convert mmHg to atm
1 atm ---------------- 760 mmHg
x --------------- 785 mmHg
x = (785 x 1) / 760
x = 1.032 atm
3.- Use the Ideal gases law
PV = nRT
solve for n
n = (PV) / (RT)
Substitution
n = (1.032 x 170) / (0.082 x 723)
Simplification
n = 175.44 / 59.29
Result
n = 2.96 moles of H₂
4.- Use the balanced reaction to calculate the moles of TiCl₄ needed
2TiCl₄ + H₂ ⇒ 2TiCl₃ + HCl
2 moles of TiCl₄ ------------- 1 mol of H₂
x -------------- 2,96 mol of H₂
x = (2.96 x 2) / 1
x = 5.92 moles of TiCl₄
5.- Convert moles to grams
190 g of TiCl₄ ---------------- 1 mol of TiCl₄
x ----------------- 5.92 moles of TiCl₄
x = (5.92 x 190) / 1
x = 1124.8 g of TiCl₄