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.
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.
0.255465
To determine the grams of ammonia produced in the reaction, we need to calculate the moles of nitrogen dioxide and water vapor, and then use their stoichiometric ratio to find the moles of ammonia produced. Finally, we can convert the moles of ammonia into grams.
Let's start by calculating the moles of nitrogen dioxide (NO2) using its molar mass:
Molar mass of NO2 = 14.01 g/mol (N) + 2 * 16.00 g/mol (O) = 46.01 g/mol
Moles of NO2 = Mass of NO2 / Molar mass of NO2
= 12.8 g / 46.01 g/mol
≈ 0.278 mol NO2
Next, we need to calculate the moles of water vapor (H2O) using the ideal gas law:
PV = nRT
Where:
P = pressure (in atm) = 725 torr = 725/760 atm
V = volume (in liters) = 5.00 L
n = moles of H2O
R = ideal gas constant = 0.0821 L.atm/mol.K
T = temperature (in Kelvin) = 375°C + 273.15 = 648.15 K
Rearranging the formula to solve for n, we have:
n = PV / RT
= (725/760 atm) * (5.00 L) / (0.0821 L.atm/mol.K) * (648.15 K)
≈ 0.186 mol H2O
Now, let's determine the stoichiometry of the reaction. From the balanced chemical equation:
2 NO2 + 4 H2O -> O2 + 4 NH3
We can see that for every 2 moles of NO2, 4 moles of NH3 are produced.
Using the stoichiometric ratio, we can find the moles of NH3 produced:
Moles of NH3 = (0.278 mol NO2) / 2 * 4
= 0.556 mol NH3
Finally, we can convert the moles of NH3 into grams by multiplying by its molar mass:
Molar mass of NH3 = 14.01 g/mol (N) + 3 * 1.01 g/mol (H) = 17.03 g/mol
Grams of NH3 = Moles of NH3 * Molar mass of NH3
= 0.556 mol * 17.03 g/mol
≈ 9.47 g of NH3
Therefore, approximately 9.47 grams of ammonia can be produced.