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

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

a) how many grams of TiCl4 are needed for complete reaction with 155L if H2 at 430 degrees C and 780mmHg pressure?

b) How any liters of HCl gas at STP will result from the reaction described in part (a)?

Respuesta :

Tofy

Answer: a. 1046.98g b. 123.64 L

Explanation:

.a. To determine the mass of TiCl4 needed to complete the reaction with 155 liters of H2 at 430 degree Celsius and 780 mmHg pressure.

First, use the ideal gas equation to determine the number of moles of H2.

PV = nRT

R= 0.082 Latm/molK

Convert Pressure to atmosphere and temperature to kelvin so the units of calculation can be uniform;

1 atm is equals to 760 mmHg so 780mmHg will be (780/760) atm = 1.0263 atm

430 degree Celsius = (430 +273.15) K

= 703.15 K

Therefore, n = PV/RT

= (1.0263*155)/ (0.082*703.15)

= 2.76 of H2 will be needed to complete the reaction

From the chemistry of the reaction, it can be seen that 2 moles of TiCl4 is needed to react with 1 mole of H2.

So for 2.76 moles of H2 , (2*2.76 moles) of TiCl4 will be needed which is equals to 5.52 moles of TiCl4.

The mass of TiCl4 can be calculated using the formular;

Mass = molar mass * number of moles

The molar mass of TiCl4 cab be calculated using the molar mass of each of its element (Ti =47.87 g/mol, Cl= 35.45 g/mol).

The molar mass of TiCl4 = 47.87 + (4*35.35) = 189.67 g/mol

Therefore,

The mass of TiCl4 = 189.67 g/mol * 5.52 mols

= 1046.98 g of TiCl4.

b. The amount of liters of HCl gas to be produced at STP. STP means standard temperature and pressure which is 273.15 K and 1 atm

From the chemistry of the reaction, it can be seen that with 1 mole of H2 is needed to produce 2 moles of HCl.

So for 2.76 moles of H2, (2*2.76 moles) of HCl will be produced which is equals to 5.52 moles of HCl. Therefore, the volume of HCl produced at 430 degree Celsius and 780 mmHg pressure will be;

V = nRT/P

= 5.52* 0.082 *703.15 / 1.0263

= 310.118 L

The combined gas law states that PV/T = constant. Therefore

P1V1/T1 = P2V2/T2

Now using the combined gas law;

P1 = 1.0263 atm, V1 = 310.118 L, T1 = 703.15 K

P2 = 1 atm, V2 =?, T2 = 273.15 k

So,

V2 = (1.0263*310.118*273.15)/ (1*703.15)

= 123.64 L

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