# At STP, if 4.20L of O_2 reacts with N_2H_4, how many liters of water vapor will be produced?

Jun 6, 2016

8.40 L of ${H}_{2} O$

#### Explanation:

I apologize in advance because this explanation is very long!

First, we need the entire chemical reaction in order to arrive at the answer:

${O}_{2}$ $+$ ${N}_{2} {H}_{4}$ $\rightarrow$ $2 {H}_{2} O$ $+$ ${N}_{2}$

For this type of problem you would use the ideal gas law equation,
$P \times V = n \times R \times T$

Where P represents pressure (must have units of atm), V represents volume (must have units of liters), n represents the number of moles, R is the proportionality constant
(has units of $\frac{L \times a t m}{m o l \times K}$), and T represents the temperature, which must be in Kelvins.

Now what you want to do is list your known and unknown variables. Our only unknown is the number of moles of water. Our known variables are P,V,R, and T. Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given volume and the proportionality constant, R, is equal to 0.0821 $\frac{L \times a t m}{m o l \times K}$.

The only problem is that we are given the volume of oxygen, so we have to convert the given volume of oxygen to moles of oxygen. Then we must use the balanced equation to go from moles of oxygen to moles of water, which will allow us to go to volume of water using the ideal gas law.

Rearrange the equation to solve for moles of ${O}_{2}$:

n $=$ $\frac{P \times V}{R \times T}$

n = (1 (cancel "atm") xx4.20(cancel"L"))/(0.0821(cancel"L"x cancel"atm")/(molxxcancel"K")xx273cancel"K") = 0.1874 mol ${O}_{2}$

Moles of $2 {H}_{2} O$ = (0.1874 (cancel ("mol${O}_{2}$) $\times$ 2 mol $2 {H}_{2} O$)/(1 (cancel "mol" ${O}_{2}$)) = 0.3748 moles ${H}_{2} O$

Rearrange the equation to solve for volume of ${H}_{2} O$:
$\frac{n \times R \times T}{P}$ $=$$V$

(0.3748 cancel"mol" xx0.0821(L xx cancel"atm")/(cancel"mol" xxcancel"K")xx273(cancel"K"))/(1cancel"atm") = 8.40 L