# Given the equation: #N_2 + 3H_2 -> 2NH_3# What volume of hydrogen is necessary to react with five liters of nitrogen to produce ammonia? (Assume constant temperature and pressure)

##### 1 Answer

#### Answer:

#### Explanation:

The important thing to remember about reactions that involve gases kept **under the same conditions for pressure and temperature** is that the mole ratios that exist between the gases are **equivalent** to *volume ratios*.

This can proven using the ideal gas law equation for two gases kept at a temperature

#P * V_1 = n_1 * R * T -># the ideal gas law equation for the first gas

#P * V_2 = n_2 * R * T -># the ideal gas law equation for the second gas

Divided these two equations to get

#(color(red)(cancel(color(black)(P))) * V_1)/(color(red)(cancel(color(black)(P))) * V_2) = (n_1 * color(red)(cancel(color(black)(R * T))))/(n_2 * color(red)(cancel(color(black)(R * T))))#

This means that you have

#color(blue)(n_1/n_2 = V_1/V_2) -># themole ratiois equivalent to thevolume ratio

In your case, the balanced chemical equation looks like this

#"N"_text(2(g]) + color(red)(3)"H"_text(2(g]) -> 2"NH"_text(3(g])#

The problem tells you that this reaction takes place at *constant temperature and pressure*. You don't need to know **the exact values**, you just need to know that they are constant for all three chemical species.

You can thus say that since you have a **volume ratio** between these two gases

#n_(N_2)/n_(H_2) = V_(N_2)/V_(H_2) = 1/color(red)(3)#

The volume of hydrogen gas needed to react with that much nitrogen gas will thus be

#5.0 color(red)(cancel(color(black)("L N"_2))) * (color(red)(3)" L H"_2)/(1 color(red)(cancel(color(black)("L N"_2)))) = color(green)("15 L H"_2)#