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

Question

Natalie Anton · Accepted Answer

8.40 L of #H_2O#

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_2H_4# #rarr# #2H_2O# #+# #N_2#

For this type of problem you would use the ideal gas law equation, #PxxV=nxxRxxT#

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 #(Lxxatm)/(molxxK)#), 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 #(Lxxatm)/(molxxK)#.

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 #=# #(PxxV)/(RxxT)#

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 #2H_2O# = (0.1874 (#cancel ("mol##O_2#)# ##xx# 2 mol #2H_2O#)/(#1 (cancel "mol" ##O_2#)) = 0.3748 moles #H_2O#

Rearrange the equation to solve for volume of #H_2O#: #(nxxRxxT)/P# #=##V#

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

Serenity Allen · Answer

undefined To find the volume of water vapor produced when 4.20 L of O2 reacts with N2H4 at STP, we first need to determine the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between O2 and N2H4 is:

$$2N_2H_4 + O_2 \rightarrow 4H_2O + N_2$$

According to the stoichiometry of the balanced equation, 1 mole of O2 reacts to produce 4 moles of water vapor.

Now, we need to calculate the number of moles of O2 present in 4.20 L of O2 at STP using the ideal gas law:

$$PV = nRT$$

At STP, P = 1 atm, T = 273 K, and R = 0.0821 L atm / (mol K). We can rearrange the equation to solve for moles:

$$n = \frac{PV}{RT}$$

Substituting the values:

$$n = \frac{(1 atm) \times (4.20 L)}{(0.0821 L \cdot \text{atm} / \text{mol} \cdot \text{K}) \times (273 K)}$$

$$n = \frac{4.20}{22.414}$$

$$n ≈ 0.1875 \text{ moles}$$

According to the stoichiometry of the balanced equation, 1 mole of O2 produces 4 moles of water vapor. Therefore, 0.1875 moles of O2 will produce:

$$0.1875 \text{ moles} \times 4 \text{ moles/L} = 0.75 \text{ moles of water vapor}$$

Using the ideal gas law again, we can find the volume of 0.75 moles of water vapor at STP:

$$V = \frac{nRT}{P}$$

$$V = \frac{(0.75 \text{ moles}) \times (0.0821 \text{ L atm} / \text{mol K}) \times (273 \text{ K})}{1 \text{ atm}}$$

$$V ≈ 16.2 \text{ L}$$

Therefore, approximately 16.2 liters of water vapor will be produced.

Axel Alvarado · Answer

undefined To answer this question, we need to consider the balanced chemical equation for the reaction between O2 and N2H4, which is:

3 O2 + 4 N2H4 -> 6 H2O + 4 N2

Using the stoichiometry of the reaction, we can see that 3 moles of O2 produce 6 moles of H2O.

Given that 4.20 L of O2 reacts, we need to calculate the volume of water vapor produced using the stoichiometry ratio.

First, convert 4.20 L of O2 to moles using the ideal gas law at STP (standard temperature and pressure, which is 0°C and 1 atm):
4.20 L O2 × (1 mol O2 / 22.4 L O2) = 0.1875 mol O2

Now, using the stoichiometry ratio from the balanced equation, we can find the moles of water produced:
0.1875 mol O2 × (6 mol H2O / 3 mol O2) = 0.375 mol H2O

Finally, convert the moles of water to liters using the ideal gas law at STP:
0.375 mol H2O × (22.4 L / mol) = 8.4 L H2O

So, 4.20 liters of O2 will produce 8.4 liters of water vapor at STP.