If 5.563 grams of #CH_4# reacted with 25.00 grams of oxygen, how much #CO_2# , in grams, can be produced? #CH_4(g) + 2O2(g) -> CO_2(g) + 2H_2O(g)#?

1 Answer
Sep 24, 2016

Answer:

Approx. thirty grams.

Explanation:

#"Moles of methane"# #=# #(5.563*g)/(16.043*g*mol^-1)# #=# #0.3468*mol#.

#"Moles of dioxygen"# #=# #(25.00*g)/(31.998*g*mol^-1)# #=# #0.7813*mol#.

Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation:

#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)#

which requires 2 equiv dioxygen per equiv of methane. So we get 1 equiv of carbon dioxide, which has a mass of #2xx0.3468*molxx44.010*g*mol^-1# #=# #??g#