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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?

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User JesseP
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1 Answer

2 votes

Answer:

1.06 × 10³ g

Step-by-step 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 * (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 * V = n * R * T\\n = (1.05atm * 155L)/((0.0821atm.L/mol.K) * 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 * (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 * (189.68g)/(mol) = 1.06 * 10^(3) g

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