chemistry problem i am stuck on

Question

Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. What volume of water would be produced by this reaction if 7.07 L of ammonia were consumed? Also, be sure your answer has a unit symbol, and is rounded to 3 significant digits.

Answer 1

Remember to write out the chemical reaction fully and make sure it is balanced.

4 NH₃ (g) + 5 O₂ (g) --> 4 NO (g) + 6 H₂O (g)

Start by converting 7.07 NH₃ to moles using the ideal gas law.

P_NH3V_NH3= n_NH3RT_NH3

P_NH3 *(7.07 L NH₃) = n_NH3* R * T_NH3

n_NH3 = P_NH3 *(7.07 L NH₃) / (R * T_NH3)

Since the NH₃ and H₂O have a 4:6 molar ratio in the reaction above, the molar quantity of H2O should be:

n_H2O = 6/4* n_NH3 = 6/4 * P_NH3 *(7.07 L NH₃) / (R * T_NH3)

Plug this expression for n_H2O into the ideal gas equation again to find the volume of H2O produced.

PV= nRT

P_H2O * V_H2O = 6/4 * P_NH3 *(7.07 L NH₃) / (R * T_NH3) * R * T_H2O

Now, for problems like these we can assume the reaction is taking place in a closed environment in which the temperature and pressure are constant. Thus, P_H2O = P_NH3 and T_H2O = T_NH3. This allows all the terms to cancel out (remember R is a constant as well), leaving:

V_H2O = 6/4 * 7.07 L NH₃ = 10.6 L

This is obviously a more tedious way to solve the question, but I wanted to show the derivation. In a reaction with gases that is carried out at constant temperature and pressure, the volumes of the gases/reactants can be substituted as the molar quantities. In other words use the volume as the number of moles.

Hope this helps, and please feel free to reach out -- happy to tutor you on additional chemistry topics.