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Elanso [62]
9 months ago
6

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

re at a temperature of 43.13 °C.
Chemistry
1 answer:
saul85 [17]9 months ago
5 0

1) List the known and unknown quantities.

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

<em>2.1- List the known and unknown quantities.</em>

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

<em>2.2- Convert grams of O2 to moles of O2.</em>

The molar mass of O2 is 31.9988 g/mol.

mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2

<em>2.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>2.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}P_{O_2}=1.57\text{ }atm

<em>The pressure of O2 is 1.57 atm.</em>

3) Find the pressure of F2.

<em>3.1- List the known and unknown quantities.</em>

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- <em>Convert grams of F2 to moles of F2.</em>

The mmolar mass of F2 is 37.9968 g/mol.

mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2

<em>3.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>3.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}P_{F_2}=1.13\text{ }atm

<em>The pressure of F2 is 1.13 atm.</em>

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

<em>4.1- Set the equation.</em>

P_T=P_A+P_B

4.2- Plug in the known quantities.

P_T=1.57\text{ }atm+1.13\text{ }atmP_T=2.7\text{ }atm

<em>The total pressure in the container is </em>2.7 atm<em>.</em>

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<h3>Answer:</h3>

64 g O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

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[Solve]   x g O₂

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                     \displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \  H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})
  2. Divide/Multiply [Cancel Units]:                                                                       \displaystyle 63.929 \ g \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

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