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

## $\text{4C"_2"H"_7"N"+13"O"_2→8"CO"+14"H"_2"O"+4"NO}$

Jun 10, 2015

The mass of $\text{CO}$ produced is 1768 grams.

The mass of $\text{H"_2"O}$ produced is 1991 grams.

#### Explanation:

Balanced equation

$\text{4C"_2"H"_7"N"+13"O"_2}$$\rightarrow$$\text{8CO"+14"H"_2"O"+4"NO}$

There is a pattern for this type of question:

mass reactant$\rightarrow$mol reactant
mol reactant$\rightarrow$mol product
mol product$\rightarrow$mass product

Mass of $\text{CO}$

1) Mass reactant$\rightarrow$ mol reactant

Convert the mass of $\text{C"_2"H"_7"N}$ to moles by dividing its given mass by its molar mass $\left(\text{45.08 g/mol}\right)$. Do this by multiplying by the inverse of its molar mass (mol/g).

2) Mol reactant to mol product

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

3) Mol product$\rightarrow$ mass product

To get the mass of $\text{CO}$, multiply mol $\text{CO}$ 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 $\text{H"_2"O}$

1) Mass reactant$\rightarrow$ mol reactant

Convert the mass of $\text{C"_2"H"_7"N}$ to moles using its molar mass by multiplying the given mass by the inverse of its molar mass $\left(\text{45.08 g/mol}\right)$.

2) Mol reactant to mol product

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

3) Mol product$\rightarrow$ mass product

To get mass of $\text{H"_2"O}$, multiply mol $\text{H"_2"O}$ 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"