Question

Titanium(III) chloride, a substance used in catalysts for preparing polyethylene, is made by high-temperature reaction of TiCl4 vapor with H2:

2 TiCl4(g) + H2(g) —> 2 TiCl3(s) + 2 HCl (g)

(a) How many grams of TiCl4 are needed for a complete reaction with 155 L of H2 at 435 °C and 795 mm Hg pressure

Answer 1

To find the grams of TiCl4 needed, we use the ideal gas law to find the moles of H2, then use stoichiometry to find the moles of TiCl4, and finally convert to grams using its molar mass.

Step-by-step explanation:

To determine the grams of TiCl4 needed for a complete reaction, we need to use the ideal gas law to find the number of moles of H2, and then use the stoichiometry of the balanced equation to find the number of moles of TiCl4. Finally, we can convert the moles of TiCl4 to grams using its molar mass.

First, let's convert the volume of H2 from liters to moles using the ideal gas law:

n(H2) = (PV) / (RT)

n(H2) = (795 mm Hg * 155 L) / (0.0821 L * atm/mol * K * 435 + 273.15)

n(H2) = 5.76 mol

Since the balanced equation shows that 2 moles of TiCl4 react with 1 mole of H2, we can use the following ratio:

2 mol TiCl4 / 1 mol H2

Finally, we can convert the moles of TiCl4 to grams using its molar mass:

mass(TiCl4) = n(TiCl4) * molar mass(TiCl4)

mass(TiCl4) = 5.76 mol * 189.7 g/mol

mass(TiCl4) ≈ 1099 g