Tricky Chemistry questions - answered by ChemTutor.me


Question:

Sample of dry gas 25c has the following composition 0.8904g of N2 0.2741g of O2 0.0152g of Ar 0.00107g of CO2 R=0.0821 What are the partial pressure for each component in the mixture?? what is the total pressure??

Answer:

Calculate the number of moles of each gas in mixture:

n(N2) = m(N2)/M(N2) = 0.8904/(14.007∙2) = 0.0317841 mol n(O2) = m(O2)/M(O2) = 0.2741/(15.999∙2) = 0.0085662 mol n(Ar) = m(Ar)/M(Ar) = 0.0152/39.948 = 0.0003805 mol

n(CO2) = m(CO2)/M(CO2) = 0.00107/(12 + 15.999∙2) = 0.0000243 mol

The total pressure of mixture cannot be found from the provided data. We can assume that the pressure of mixture is atmospheric. Hence we can calculate the partial pressure of each component in mixture.

n(total) = 0.31784 + 0.008566 + 0.0003805 + 0.0000243 = 0.040755 mol

x(N2) = n(N2)/n(total) = 0.7799

x(O2) = n(O2)/n(total) = 0.2102

x(Ar) = n(Ar)/n(total) = 0.00934

x(CO2) = n(CO2)/n(total) = 0.000596

pi = xi∙p(total)

p(N2) = n(N2)/n(total) = 0.7799∙101325 Pa = 79020 Pa p(O2) = n(O2)/n(total) = 0.2102∙101325 Pa = 21300 Pa p(Ar) = n(Ar)/n(total) = 0.00934∙101325 Pa = 946 Pa

p(CO2) = n(CO2)/n(total) = 0.000596∙101325 Pa = 60.4 Pa


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