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laiz [17]
3 years ago
13

Oxygen gas was produced in a reaction and collected over water. A 136.1 mL mL sample of gas was collected over water at 25C and

1.06 atm. The vapor pressure of water is 23.76 mmHg at 25C. Find the mass of ocygen gas collected in a reaction and collected over water. A 136.1 mL sample of gas was collected over water at 25C and 1.06 atm. The vapor pressure of water is 23.76 mmHg at 25C. Find the mass of oxygen gas collected.
Chemistry
1 answer:
Xelga [282]3 years ago
4 0

Answer:

Explanation:

We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .

Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .

= 805.6 - 23.76 = 781.84 mm of Hg

For it we use gas law formula ,

P₁V₁ / T₁ = P₂V₂ / T₂

781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273

= 128.26 mL .

= 128.26  x 10⁻³ L .

22.4 L of oxygen will have mass of 32 g

128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g

= 183.22 mg .

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Answer: 2 moles of H_2O will be formed.

Explanation:

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\text{Moles of} NaOH=\frac{80g}{40g/mol}=2moles

The balanced chemical equation is:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to stoichiometry :

2 moles of NaOH give = 2 moles of H_2O

Thus 2 moles of NaOH give =\frac{2}{2}\times 2=2moles of H_2O

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A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3
Zolol [24]

Answer:

The mass of NH_{3} in the container is 2.074 gram

Explanation:

Given:

Volume of NH_{4} HS    V = 2 lit

Equilibrium constant k _{eq} = 3.5 \times 10^{-3}

The reaction in which NH_{3} is produced

  NH_{4} HS ⇄ NH_{3} + H_{2}S

Here equal moles of NH_{3} and H_{2}S is formed.

From the formula of equilibrium constant,

  k_{eq} = (NH_{3})(H_{2}S )

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