# Consider the following balanced equation. 2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g) ?

## Complete the following table, showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product formed. If the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that is made.

Feb 23, 2018

Warning! Long Answer. Here's what I get.

#### Explanation:

$\underline{\boldsymbol{\text{Compd. ="color(white)(m)"N"_2"H"_4color(white)(m)"N"_2"O"_4color(white)(m)"N"_2color(white)(m)"H"_2"O}}}$
$\textcolor{w h i t e}{m m m l l} n = \textcolor{w h i t e}{m m l} 4 \textcolor{w h i t e}{m m m} \textcolor{red}{2} \textcolor{w h i t e}{m m m} 6 \textcolor{w h i t e}{m m l l} 8$
$\textcolor{w h i t e}{m m m l l} n = \textcolor{w h i t e}{m m l} \textcolor{red}{2} \textcolor{w h i t e}{m m m} 1 \textcolor{w h i t e}{m m m} 3 \textcolor{w h i t e}{m m l l} 4$
$\textcolor{w h i t e}{m m m l l} n = \textcolor{w h i t e}{m m l} 4 \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m m} 6 \textcolor{w h i t e}{m m l l} \textcolor{red}{8}$

$\textcolor{w h i t e}{m m m l l} n = \textcolor{w h i t e}{m m l} 4.0 \textcolor{w h i t e}{m m} \textcolor{red}{2.0} \textcolor{w h i t e}{m m l} 6.0 \textcolor{w h i t e}{m l l} 8.0$
$\textcolor{w h i t e}{m m m l l} n = \textcolor{w h i t e}{m m l} \textcolor{red}{1.4} \textcolor{w h i t e}{m m} 0.70 \textcolor{w h i t e}{m l l} 2.1 \textcolor{w h i t e}{m l l} 2.8$
$\textcolor{w h i t e}{m m m l l} n = \textcolor{w h i t e}{m l l} 22.2 \textcolor{w h i t e}{m l l} 11.1 \textcolor{w h i t e}{m l l} \textcolor{red}{33.3} \textcolor{w h i t e}{m} 44.4$

This question involves practice in using molar ratios.

The balanced equation is

$\text{2N"_2"H"_4 + "N"_2"O"_4 → "3N"_2 +"4H"_2"O}$

Row 1

${\text{Moles of N"_2"H"_4 = 2 color(red)(cancel(color(black)("mol N"_2"O"_4))) × ("2 mol N"_2"H"_4)/(1 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "4 mol N"_2"H}}_{4}$

${\text{Moles of N"_2 = 2 color(red)(cancel(color(black)("mol N"_2"O"_4))) × ("3 mol N"_2)/(1 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "6 mol N}}_{2}$

$\text{Moles of H"_2"O" = 2 color(red)(cancel(color(black)("mol N"_2"O"_4))) × ("4 mol H"_2"O")/( 1 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "8 mol H"_2"O}$

Do you notice a pattern?

You are using twice as many moles of ${\text{N"_2"O}}_{4}$ as in the balanced equation, so you must use twice as many moles of everything else.

The conversion factor is always the molar ratio: $\text{moles of what you want"/"moles of what you have}$.

Row 2

${\text{Moles of N"_2"O"_4 = 2 color(red)(cancel(color(black)("mol N"_2"H"_4))) × ("1 mol N"_2"H"_4)/(2 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "1 mol N"_2"H}}_{4}$

You are using the same number of moles of ${\text{N"_2"O}}_{4}$ as in the balanced equation, so you must use the same amount of everything else as in the balanced equation.

You can do the calculation using numbers as above, and you will get ${\text{3 mol N}}_{2}$ and $\text{4 mol H"_2"O}$.

Row 3

${\text{Moles of N"_2"H"_4 = 8 color(red)(cancel(color(black)("mol H"_2"O"))) × ("2 mol N"_2"H"_4)/(4 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "4 mol N"_2"H}}_{4}$

You are using twice as many moles of ${\text{H"_2"O}}_{4}$ as in the balanced equation, so you must use twice as many moles of everything else.

Thus, you will get ${\text{2 mol N"_2"O}}_{4}$ and ${\text{6 mol N}}_{2}$.

Row 4

Row 4 is the same as Row 2, but you are using more precision in your measurements (one more decimal point).

Thus, you get ${\text{4.0 mol N"_2"H"_4, "6.0 mol N}}_{2}$, and $\text{8.0 mol H"_2"O}$.

Row 5

${\text{Moles of N"_2"O"_4 = 1.4 color(red)(cancel(color(black)("mol N"_2"H"_4))) × ("1 mol N"_2"H"_4)/(2 color(red)(cancel(color(black)("mol N"_2"O"_4)))) = "0.70 mol N"_2"H}}_{4}$

You are using 0.70 times as many moles of ${\text{N"_2"H}}_{4}$ as in the balanced equation, so you must use 0.70 times as many moles of everything else.

Thus, you will get $\text{2.1 mol N"_2}$ and $\text{2.8 mol H"_2"O}$.

Row 6

${\text{Moles of N"_2"H"_4 = 33.3 color(red)(cancel(color(black)("mol N"_2))) × ("2 mol N"_2"H"_4)/(3 color(red)(cancel(color(black)("mol N"_2)))) = "22.2 mol N"_2"H}}_{4}$

You are using 11.1 times as many moles of ${\text{N}}_{2}$ as in the balanced equation, so you must use 11.1 times as many moles of everything else.

Thus, you will get ${\text{11.1 mol N"_2"O}}_{4}$ and $\text{44.4 mol H"_2"O}$.