#"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"#

Answer 1

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

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

Equation in balance

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

This kind of question follows a pattern:

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).
  1. Mol product to mol reactant
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")#.
  1. Mol product to mol reactant
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"#
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Answer 2

To determine the mass of CO and H2O produced, we need to know the extent of incomplete combustion, typically represented by a mole ratio or percentage yield. Without that information, it's impossible to calculate the exact mass of CO and H2O produced.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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