#"1423 g C"_2"H"_7"N"# undergoes incomplete combustion. What mass of #"CO"# and #"H"_2"O"# can be produced? The equation is:

#"4C"_2"H"_7"N"+13"O"_2→8"CO"+14"H"_2"O"+4"NO"#

1 Answer
Jun 10, 2015

Answer:

The mass of #"CO"# produced is 1768 grams.

The mass of #"H"_2"O"# produced is 1991 grams.

Explanation:

Balanced equation

#"4C"_2"H"_7"N"+13"O"_2"##rarr##"8CO"+14"H"_2"O"+4"NO"#

There is a pattern for this type of question:

mass reactant#rarr#mol reactant
mol reactant#rarr#mol product
mol product#rarr#mass product

Mass of #"CO"#

1) Mass reactant#rarr# mol reactant

Convert the mass of #"C"_2"H"_7"N"# to moles by dividing its given mass by its molar mass #("45.08 g/mol")#. Do this by multiplying by the inverse of its molar mass (mol/g).

2) Mol reactant to mol product

To get mol #"CO"#, multiply mol #"C"_2"H"_7"N"# by the mole ratio between #"CO"# and #"C"_2"H"_7"N"# in the balanced equation, with #"CO"# in the numerator.

3) Mol product#rarr# mass product

To get the mass of #"CO"#, multiply mol #"CO"# by its molar mass.

#1423color(red)cancel(color(black)("g C"_2"H"_7"N"))xx(1color(red)cancel(color(black)("mol C"_2"H"_7"N")))/(45.08color(red)cancel(color(black)("g C"_2"H"_7"N")))xx(8color(red)cancel(color(black)("mol CO")))/(4color(red)cancel(color(black)("mol C"_2"H"_7"N")))xx(28.01"g CO")/(1color(red)cancel(color(black)("mol CO")))=1768"g CO"#

Mass of #"H"_2"O"#

1) Mass reactant#rarr# mol reactant

Convert the mass of #"C"_2"H"_7"N"# to moles using its molar mass by multiplying the given mass by the inverse of its molar mass #("45.08 g/mol")#.

2) Mol reactant to mol product

To get mol #"H"_2"O"#, multiply mol #"C"_2"H"_7"N"# by the mole ratio between #"H"_2"O"# and #"C"_2"H"_7"N"# in the balanced equation, with #"H"_2"O"# in the numerator.

3) Mol product#rarr# mass product

To get mass of #"H"_2"O"#, multiply mol #"H"_2"O"# by its molar mass.

#1423color(red)cancel(color(black)("g C"_2"H"_7"N"))xx(1color(red)cancel(color(black)("mol C"_2"H"_7"N")))/(45.08color(red)cancel(color(black)("g C"_2"H"_7"N")))xx(14color(red)cancel(color(black)("mol H"_2"O")))/(4color(red)cancel(color(black)("mol C"_2"H"_7"N")))xx(18.02"g H"_2"O")/(1color(red)cancel(color(black)("mol H"_2"O")))=1991"g H"_2"O"#