How many grams of ticl4 are needed for complete reaction with 170 l of h2 at 450 ∘c and 785 mm hg pressure?

The answer to your question is 1124.8 g of TiCl₄

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

it means there are 3 atoms of oxygen