Nitrogen dioxide abstracts with water vapor to produce oxygen and ammonia gas. If 12.8 grams of nitrogen dioxide reacts with 5.00 liter of water vapor at 375 degrees celisus and 725 tore, how many grams of ammonia can be produced? Use an ice table.

Answer 1

This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.

4NO2 + 6H2O ==> 4NH3 + 7O2

mols NO2 = grams/molar mass = ?
mols H2O = use PV = nRT and solve for n.
Using the coefficients in the balanced equation, convert mols NO2 to mols NH3.
Do the same for mols H2O to mols NH3.
It is likely that the values for mols NH3 will not agree which means one of them is incorrect; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Then grams NH3 = smaller value of mols x molar mass NH3 = ?
Post your work if you get stuck.

Answer 2

0.255465