The following reaction takes place in a sealed 40.0-L container at a temp of 120 degrees C
4NH 3(G) + 5 O2 (g) --> 4NO (g) + 6H2O (G)
a) When 34.0g of NH3 reacts with 96.0g of O2 what is the partial pressure of NO in the sealed container?
b) What is the total pressure in the container?
First change grams to moles...
34g NH3 (1 mol/ 17 g) = 2.00 mol NH3
96g O2 (1 mol/32 g) = 3.00 mol O2
This is a limiting reactant problem.
To figure out how much O2 is required to react w/ 2 mol NH3...
2 mol NH3[(5 mol O2)/(4 mol NH3)] = 2.5 mol O2
However, we have 3 mol O2... so there is more than enough O2, so that means that NH3 is the limiting ...
This solution is provided in approximately 327 words. It uses conversion of grams to moles and the ideal gas law to find total and partial pressure, as well as discusses limiting reactants in the reaction.