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Naddik [55]
3 years ago
14

2C6H10 + 17O2 = 12CO2 + 10H2O

Chemistry
1 answer:
Ad libitum [116K]3 years ago
6 0

Answer:

n O2 = 2.125 mol

Explanation:

balanced reaction:

  • 2C6H10 + 17O2 → 12CO2 + 10H2O

∴ n CO2 = 1.5 mol

⇒ n O2 = (1.5 mol CO2)*(17 mol O2/12 mol CO2)

⇒ n O2 = 2.125 mol

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24
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<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

  • <u>For Part A:</u>

We are given:

p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg

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  • <u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

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PV=\frac{m}{M}RT

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P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g

Hence, the mass of helium gas is 0.504 g.

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