A compound with molecular formula #CH_4O# burns in air to form carbon dioxide and water. How do you write a balanced equation for this reaction?

Question

A compound with molecular formula #CH_4O# burns in air to form carbon dioxide and water.  How do you write a balanced equation for this reaction?

Olivia Anton · Accepted Answer

undefined CH₄O + 2O₂ → CO₂ + 2H₂O

Avery Adams · Answer

Warning! Long answer! The balanced equation is
#"2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O"# To balance the equation, you adhere to a methodical process. To begin, consider the imbalanced equation: #"CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O"# A method that often works is to balance everything other than #"O"# and #"H"# first, then balance #"O"#, and finally balance #"H"#. Starting with the formula that appears to be the most complex is another helpful approach. The most complicated formula looks like #"CH"_4"O"#. We put a #1# in front of it to remind ourselves that the coefficient is now fixed. We begin with #color(red)(1)"CH"_4"O" + "O"_2 → "CO"_2 + "H"_2"O"# Balance #"C"#: We have fixed 1 #"C"# atom on the left-hand side, so we need 1 #"C"# atom on the right-hand side. We put a #1# in front of the #"CO"_2#. #color(red)(1)"CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + "H"_2"O"# Balance #"O"#: We can't balance #"O"# yet, because we have two formulas that contain #"O"# and lack coefficients. So we balance #"H"# instead. Balance #"H"#: We have fixed 4 #"H"# atoms on the left-hand side, so we need 4 #"H"# atoms on the right-hand side. We put an #2# in front of the #"H"_2"O"#. #color(red)(1)"CH"_4"O" + "O"_2 → color(blue)(1)"CO"_2 + color(green)(2)"H"_2"O"# Now we balance #"O"#: We have fixed 4 #"O"# atoms on the right-hand side: 2 from the #"CO"_2# and 2 from the #"H"_2"O"#. We have fixed 1 #"O"# atom on the left-hand side, so we need 3 more from the #"O"_2#. Oops. We would need a fraction of an #"O"_2# molecule. Restarting, we multiply each coefficient by two. #color(red)(2)"CH"_4"O" + "O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O"# Now we have fixed 8 #"O"# atoms on the right and 2 on the left. We need 6 more #"O"# atoms on the left. We put a 3 in front of the #"O"_2#. #color(red)(2)"CH"_4"O" + color(brown)(3)"O"_2 → color(blue)(2)"CO"_2 + color(green)(4)"H"_2"O"# Now that each formula has a fixed coefficient, the equation ought to be balanced. Let us verify: #mathbf(color(white)(m)"Element"color(white)(m)"Left-hand side"color(white)(m)"Right-hand side")# #color(white)(mmll)"C"color(white)(mmmmmml)2color(white)(mmmmmmmmll)2# #color(white)(mmll)"H"color(white)(mmmmmml)8color(white)(mmmmmmmmll)8# #color(white)(mmll)"O"color(white)(mmmmmml)8color(white)(mmmmmmmmll)8# Atoms all balance. The equation that is in balance is #"2CH"_4"O" + "3O"_2 → "2CO"_2 + "4H"_2"O"#