59 ) First calculate gm of each gas Cyclopropane Use ideal gas equation for calc

Question

59 ) First calculate gm of each gas

Cyclopropane

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 0.720 atm,

V = volume in Liter = 0.450 L

n = number of mole = ?

R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,

T = Temperature in K = 1200C = 273.15+ 120 = 393.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.720X 0.450)/(0.08205 X 393.15) = 0.00937 mole

molar mass of cylcopropane = 42.08 gm/mole then 0.00937 mole of cyclopropane = 0.00937 X 42.08 = 0.394 gm

gm of cyclopropane = 0.394 gm

CO2

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 0.720 atm,

V = volume in Liter = 1.35 L

n = number of mole = ?

R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,

T = Temperature in K = 1200C = 273.15+ 120 = 393.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.720X 1.35)/(0.08205 X 393.15) = 0.03 mole

molar mass of CO2 = 44.01 gm/mole then 0.03 mole of CO2= 0.03 X 44.01 = 1.32 gm

gm of CO2= 1.32 gm

Molar mass of CO2 = 44.01 gm/mole and molar mass of C = 12.0107 gm/mole that mean 44.01 gm of CO2 contain 12.0107 gm of C then 1.32 gm of CO2 contain C = 1.32 X 12.0107 / 44.01 = 0.36 gm C

0.394 gm cyclopropane = 100 % then 0.362 gm C = 0.362 X 100 / 0.394 = 91.4%

91.4% of carbon

Explanation / Answer

59. Cyclopropane, a gas containing only carbon and hydrogen, is an anesthetic If 0.450 L of cyclopropane at 120 °C and 0.720 atm reacts with excess 02 to give 1.35 L of CO2(g) and 1.35 L of H20(g) at the same temperature and pressure, what is the percent by weight of carbon in cyclopropane?

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.