1 Normal No Spac... Heading 1 Head Paragraph Sty 27. EXTRA CREDIT! If you add 36

Question

1 Normal No Spac... Heading 1 Head Paragraph Sty 27. EXTRA CREDIT! If you add 36.0 ml of CHs (methane) gas at STP to a 1.43 L container, seal the container, and combust the gas with 18.50 mL of oxygen. (Molar volume of a gas at STP-22.4 LImol a. What is the balanced chemical equation for the reaction20) b. What is the limiting reagent? What are the theoretical moles of CO2 produced? (3) c. Use Dalton's Law of Partial Pressures and the ideal gases law (PV- nRT to determine the partial pressures of each of the gases when the reaction has gone to completion at 298.6 K? (5)

Explanation / Answer

Q1

a

CH4 + O2 = CO2 + H2O

balance

CH4 + 2O2 = CO2 + 2H2O

b)

limiting:

36 mL CH4 ; 18.5 mL of O2

clearly there is excess CH4

ratio is 2:1 so

18.5 mL of O2 = 1/2*18.5 = 9.25 mL of CO2 expected

1 mol = 22400 mL

x = 9.25 mL

x = 9.25/22400

x = 0.0004129 mol of CO2

c)

mol of CH4 = 36/22400 = 0.0016071

mol of CH4 afte reaction = 0.0016071 - 0.0004129 = 0.0011942

mol of O2 left = 0

mol of CO2 = 0.0004129  

mol of H2O = 2*0.0004129 = 0.0008258

Total n = 0.0011942+0.0004129 +0.0008258 = 0.0024329

now..

P = nRT/V = (0.0024329*0.082)(298.6)/1.43 =0.0416573 atm

P-CH4 = mol of CH4/ total mol * Ptotal = 0.0011942/0.0024329*0.0416573 = 0.02044 atm

P-CO2= mol of CO2/ total mol * Ptotal = 0.0004129  /0.0024329*0.0416573 = 0.007069 atm

P-H2O = mol of H2O / total mol * Ptotal = 0.0008258/0.0024329*0.0416573 = 0.014139 atm

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