How many moles of ethylene (#C_2H_4#) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin?

1 Answer
Jul 9, 2016

Answer:

#0.212 \ mol. C_2H_4#

Explanation:

Start by using the ideal gas equation to figure out the the number of moles #O_2# used.

# P*V= n*R*T#

Where

#P" "# is the pressure expressed in atm.

#V" "# is the volume expressed in L.

#n" " # is the number of moles.

#R" "# is the universal gas constant, it has a value of #0.0821\ L* atm. mol^-1*K^-1#

#T" " #is the Kelvin temperature.

Now rearrange the ideal gas equation and solve for n.

#n = (P*V)/(R*T)#

#n = (12.9 \ Lxx1.2 \ atm)/(0.0821\ L* atm. mol^-1*K^-1 xx 297 \ K)#

#n = (12.9 \ cancel(L)xx1.2 \ cancel( atm))/(0.0821\ cancel(L) *cancel(atm) *mol^-1*cancel(K^-1) xx 297 \ cancel(K))#

#n = 0.635 \ mol.#

In the second step, write a balanced chemical equation for the reaction and use the stoichiometry of the equation to figure out the moles of ethylene used.

#C_2H_4 + 3O_2 -> 2CO_2 + 2H_2O#

#0.635 \ mol. O_2 xx (1\ mol. \ C_2H_4)/(3 \ mol. \ O_2) #

#0.635 \ cancel (mol. O_2) xx (1\ mol. \ C_2H_4)/(3 \ cancel( mol. \ O_2))#

#0.212 \ mol. C_2H_4#