Question

g A mixture of gases contains 6.46 g of N2O, 2.74 g of CO, and 5.40 g of O2. If the total pressure of the mixture is 4.33 atm, what is the partial pressure of each component? a) P(N2O) = 0.635 atm, P(CO) = 0.424 atm, and P(O2) = 3.27 atm. b) P(N2O) = 2.31 atm, P(CO) = 0.622 atm, and P(O2) = 1.40 atm. c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm. d) P(N2O) = 0.999 atm, P(CO) = 0.371 atm, and P(O2) = 2.96 atm. e) P(N2O) = 1.28 atm, P(CO) = 1.93 atm, and P(O2) = 1.12 atm.

Answer 1

The correct answer is c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm

Step-by-step explanation:

In order to calculate the partial pressures of the mixture components, we have to first calculate the number of moles:

For N₂O:

Molecular weight (MW): (14 g/mol x 2) + 16 g/mol= 44 g/mol

Number of moles of N₂O (n) = mass/Mw = 6.46g/44 g/mol= 0.1468 mol

For CO:

Molecular weight (MW): 12 g/mol + 16 g/mol= 28 g/mol

Number of moles of CO (n) = mass/Mw = 2.74 g/28 g/mol= 0.0978 mol

For O₂:

Molecular weight (MW): 16 g/mol x 2= 32 g/mol

Number of moles of O₂ (n) = mass/Mw = 5.40 g/32 g/mol= 0.1687 mol

Once calculated the number of moles of each component, we can calculate the total number of moles (nt):

nt = 0.1468 mol + 0.0978 mol + 0.1687 mol = 0.4133 moles

The partial pressure of a gas in a mixture can be calculated from the molar fraction of the gas (X) and the total pressure of the mixture (Pt=4.33 atm):

P(N₂O) = X(N₂O) x Pt

= (moles N₂O/nt) x Pt

= 0.1468 moles/0.4133 moles x 4.33 atm

= 1.538 atm

P(CO) = X(CO) x Pt

= (moles CO/nt) x Pt

= 0.0978 moles/0.4133 moles x 4.33 atm

= 1.0246 atm

P(O₂) = X(O₂) x Pt

= (moles O₂)/nt x Pt

= 0.1687 moles/0.4133 moles x 4.33 atm

= 1.767 atm