Mole fractions Part A What is the mole fraction of O2 in a mixture of 15.1g of O

Question

Mole fractions

Part A

What is the mole fraction of O2 in a mixture of 15.1g of O2, 8.19g of N2, and 2.47g of H2?

Part B

What is the mole fraction of N2 in a mixture of 15.1g of O2, 8.19g of N2, and 2.47g of H2?

Part C

What is the mole fraction of H2 in a mixture of 15.1g of O2, 8.19g of N2, and 2.47g of H2?

Part D

What is the partial pressure in atm of O2 of this mixture if it is held in a 15.50?L vessel at 14?C?

Part E

What is the partial pressure in atm of N2 of this mixture if it is held in a 15.50?L vessel at 14?C?

Part F

What is the partial pressure in atm of H2 of this mixture if it is held in a 15.50?L vessel at 14?C?

Explanation / Answer

Part A

molar masses of H2, O2 and N2 are respectively 2, 32 and 28

moles of H2 = 2.47 / 2 = 1.235

moles of O2 = 15.1 /32 = 0.472

moles of N2 = 8.19 / 28 = 0.2925

TOTAL MOLES = 1.235 + 0.472 + 0.2925 = 1.9995

Mole fraction of O2 = moles of O2/ total moles = 0.472 / 1.9995

                            = 0.236

Part B

Mole fraction of N2 = moles of N2/ total moles = 0.2925 / 1.9995

                            = 0.146

Part C

Mole fraction of H2 = moles of H2/ total moles = 1.235 / 1.9995

                            = 0.618

Part D

V = 15.5 L

T = 14 C + 273 = 287 K

n = 0.472

P = nRT / V = 0.472 * 0.0821 * 287 / 15.5

    = 0.72 atm

Part E

V = 15.5 L

T = 14 C + 273 = 287 K

n = 0.2925

P = nRT / V = 0.2925 * 0.0821 * 287 / 15.5

    = 0.44 atm

Part D

V = 15.5 L

T = 14 C + 273 = 287 K

n = 1.235

P = nRT / V = 1.235 * 0.0821 * 287 / 15.5

    = 1.88 atm

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