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kaheart [24]
3 years ago
8

A 10 liter flask at 298 K contains a gaseous mixture of O2 and CO2 at 1 atmosphere. Which statement is true for the partial pres

sures of O2 and CO2 if 0.2 mole of O2 is present in the flask? (Given the universal gas constant R = 0.082 L∙atm/K∙mol)
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

The partial pressures of O₂ and CO₂ are 0.489 atm and 0.511 atm respectively.

Explanation:

From the Questions we are given;

Volume = 10 Liter

Temperature = 298 K

Pressure = 1 atm

We need to calculate the partial pressures of O₂ and CO₂

Step 1 : Number of moles of gaseous mixture

Using the ideal gas equation;

PV =nRT, where P is the pressure, V is the volume, n is the number of moles, T is the temperature in K and R is the ideal gas constant (0.082 L∙atm/K∙mol)

Therefore;

n =\frac{PV}{RT}

n = \frac{(10)(1)}{(298)(0.082)}

Solving for n

n = 0.409 moles

Step 2: Moles of CO₂

Total number of moles of the mixture = 0.409 moles

Moles of Oxygen = 0.2 moles

Therefore;

Moles of CO₂ = 0.409 moles - 0.2 moles

                      = 0.209 moles

Step 3: Partial pressures of O₂ and CO₂

Partial pressure = \frac{No. of moles}{Total number of moles}(total pressure)

Therefore;

Partial pressure of Oxygen gas

= \frac{number of moles of Oxygen}{Total number of moles} (Total pressure)

= \frac{0.2}{0.409}(1)\\= 0.489 atm

Partial pressure of CO₂

= \frac{number of moles of CO₂}{Total number of moles} (Total pressure)

= \frac{0.209}{0.409}(1)\\= 0.511 atm

Thus, the partial pressures of O₂ and CO₂ are 0.489 atm and 0.511 atm respectively.

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Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4
vova2212 [387]

Answer:

a) 226.6 grams of Cl₂

b) 19.2 grams of C

c) 303.2 grams of TiCl₄ and 70.4 grams of CO₂

Explanation:

The balanced chemical reaction is the following:

TiO₂(s) + C(s) + 2 Cl₂(g) → TiCl₄(s) + CO₂(g)

(a) What mass of Cl₂ gas is needed to react with 1.60 mol TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 2 moles of Cl₂. So, the stoichiometric ratio is 2 mol Cl₂/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of Cl₂ we need:

1.60 mol TiO₂ x 2 mol Cl₂/1 mol TiO₂ = 3.2 mol Cl₂

Now, we convert from moles to mass by using the molecular weight (MW) of Cl₂:

MW(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

mass of Cl₂= 3.2 mol x 70.8 g/mol = 226.6 g

<em>Therefore, 226.6 grams of Cl₂ are needed to react with 1.6 mol of TiO₂. </em>

(b) What mass of C is needed to react with 1.60 mol of TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 1 moles of C(s). So, the stoichiometric ratio is 1 mol C/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of C(s) we need:

1.60 mol TiO₂ x 1 mol C(s)/1 mol TiO₂ = 1.60 mol C(s)

So, we convert the moles of C(s) to grams as follows:

MW(C) = 12 g/mol

1.60 mol x 12 g/mol = 19.2 g C(s)

<em>Therefore, a mass of 19.2 grams of C is needed to react with 1.60 mol of TiO₂. </em>

(c) What is the mass of all the products formed by reaction with 1.60 mol of TiO₂?

From the chemical equation, we can notice that 1 mol of TiO₂ produces 1 mol of TiCl₄ and 1 mol of CO₂. So, from 1.60 moles of TiO₂, 1 mol of each product will be produced:

1 mol TiO₂/1 mol TiCl₄ ⇒ 1.60 mol TiO₂/1.60 mol TiCl₄

1 mol TiO₂/1 mol CO₂ ⇒ 1.60 mol TiO₂/1.60 mol CO₂

Finally, we convert the moles to grams by using the molecular weight of each compound:

MW(TiCl₄) = 47.9 g/mol Ti + (35.4 g/mol x 4 Cl) = 189.5 g/mol

1.60 mol x 189.5 g/mol = 303.2 g

MW(CO₂) = 12 g/mol C + (16 g/mol x 2 O) = 44 g/mol

1.60 mol x 44 g/mol = 70.4 g

<em>Therefore, from the reaction of 1.60 mol of TiO₂ are formed 303.2 grams of TiCl₄ and 70.4 grams of CO₂.</em>

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