What mass of #"CO"_2"# will be produced when #"17.48 g CO"# reacts with excess #"O"_2"#?

1 Answer
Aug 23, 2017

Answer:

Under the conditions given, #"27.46 g CO"_2# will be produced.

Explanation:

Balanced Equation

#"2CO(g) + O"_2("g")##rarr##"2CO"_2("g")"#

First, determine mol #"CO"# in #"17.48 g"# by dividing its given mass by its molar mass #("28.010 g/mol")#. Since molar mass is a fraction, g/mol , you can multiply the given mass by the inverse of the molar mass, mol/g .

Next multiply the moles #"CO"# by the mole ratio between #"CO"# and #"CO"_2"# from the balanced equation with #"CO"_2# in the numerator.

Finally, multiply mol #"CO"_2"# by its molar mass #("44.009 g/mol")#.

#17.48color(red)cancel(color(black)("g CO"))xx(1color(red)cancel(color(black)("mol CO")))/(28.010color(red)cancel(color(black)("g CO")))xx(2color(red)cancel(color(black)("mol CO"_2)))/(2color(red)cancel(color(black)("mol CO")))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))="27.46 g CO"_2"# (rounded to four significant figures)