If #11# grams of carbon dioxide is produced, how much methane, #"CH"_4#, must have reacted initially in moles?

#"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(l)#

2 Answers

Answer:

From the balanced equation determine the ratio of moles of carbon dioxide produced to methane reacted, then from the molar mass of carbon dioxide get the moles of carbon dioxide and multiply by the methane to #"CO"_2# ratio.

Explanation:

The balanced equation shows that #1.0# mole of #"CH"_4# reacts with excess oxygen to form #1.0# mole of #"CO"_2#.

Calculating the molar mass of #"CO"_2#, we see that #"C"# is #"12 g/mol"# and #"O"# is #"16 g/mol"#, so #"CO"_2# has a #M_M# of #"44 g/mol"#.

Dividing #"11 g"# #"CO"_2# by #"44 g/mol"# #"CO"_2# gives #0.25# mol #"CO"_2#. The #1:1# ratio of methane to #"CO"_2# means that #0.25# mol #"CH"_4# originally reacted.

Answer:

The answer has to be #0.25# mol.

Explanation:

Methane undergoes a combustion reaction as follow:

#"CH"_4 +2"O"_2 -> "CO"_2 + 2"H"_2"O"#

#n_("CO"_2)=m/M_r = "11 g"/((12.01+(2xx16)) \ "g/mol") = "0.25 mol"#

According to the balanced equation, #1# mole of #"CH"_4# reacts with #2# moles of #"O"_2# to produce #1# mole of #"CO"_2#. That means

#"1 mol" ("CO"_2) = "1 mol" ("CH"_4)#

Therefore

#n_("CH"_4) = "0.25 mol"#