4.00kg nitrogen dioxide reacts with 1.00kg water to produce nitric acid and nitrogen monoxide. Determine the limiting reactant and the theoretical yield of nitric acid in grams. How many grams of excess reactant are present after the reaction is complete?
To determine the limiting reactant and the theoretical yield of nitric acid, we need to compare the amount of product that each reactant can produce.
1. Calculate the moles of each reactant:
Molar mass of nitrogen dioxide (NO2) = 46.01 g/mol
Molar mass of water (H2O) = 18.02 g/mol
Moles of nitrogen dioxide = mass / molar mass = 4.00 kg / 46.01 g/mol = 86.90 mol
Moles of water = mass / molar mass = 1.00 kg / 18.02 g/mol = 55.49 mol
2. Write the balanced chemical equation for the reaction:
2 NO2 + H2O -> HNO3 + NO
According to the balanced equation, for every 2 moles of nitrogen dioxide, we need 1 mole of water to produce 1 mole of nitric acid.
3. Determine the limiting reactant:
The ratio of moles of nitrogen dioxide to moles of water is 2:1.
Since we have 86.90 moles of nitrogen dioxide and 55.49 moles of water, we can see that there is an excess of nitrogen dioxide. Therefore, the limiting reactant is water.
4. Calculate the theoretical yield of nitric acid:
From the balanced equation, we can see that 2 moles of NO2 react to produce 1 mole of HNO3.
Moles of NO2 used = moles of water * (2 moles NO2 / 1 mole H2O) = 55.49 mol * 2/1 = 110.98 mol
Theoretical yield of nitric acid = moles of NO2 used * molar mass of HNO3
= 110.98 mol * 63.01 g/mol = 6,998.31 g
Therefore, the theoretical yield of nitric acid is 6,998.31 grams.
5. Calculate the grams of excess reactant:
Excess moles of nitrogen dioxide = moles of nitrogen dioxide - moles of NO2 used = 86.90 mol - 110.98 mol = -24.08 mol
The negative value indicates that there is no excess nitrogen dioxide left. So, there are 0 grams of excess nitrogen dioxide present after the reaction is complete.
To summarize:
- The limiting reactant is water.
- The theoretical yield of nitric acid is 6,998.31 grams.
- There is no excess nitrogen dioxide present after the reaction is complete.
To determine the limiting reactant and theoretical yield, we need to compare the given masses of nitrogen dioxide (NO2) and water (H2O) to the stoichiometry of the balanced chemical equation. The balanced equation for the reaction is as follows:
2NO2 + H2O -> HNO3 + NO
The molar mass of NO2 is 46.01 g/mol, and the molar mass of H2O is 18.02 g/mol. To convert the given masses of NO2 and H2O to moles, we use the following equations:
Moles of NO2 = Mass of NO2 / Molar mass of NO2
Moles of H2O = Mass of H2O / Molar mass of H2O
Moles of NO2 = 4.00 kg / (46.01 g/mol) = 86.91 mol
Moles of H2O = 1.00 kg / (18.02 g/mol) = 55.50 mol
According to the balanced equation, the ratio of NO2 to H2O should be 2:1. Since the ratio is not equal, one of the reactants will be completely consumed, making it the limiting reactant. To determine which reactant is limiting, we need to calculate the amount of HNO3 produced by each reactant.
To calculate the moles of HNO3 produced, we use the stoichiometry of the balanced equation:
Moles of HNO3 = 0.5 x Moles of NO2
Moles of HNO3 = 0.5 x 86.91 mol = 43.45 mol
Moles of HNO3 = 1 x Moles of H2O
Moles of HNO3 = 1 x 55.50 mol = 55.50 mol
From the calculations, we can see that the moles of HNO3 produced by NO2 (43.45 mol) are less than the moles produced by H2O (55.50 mol). Therefore, NO2 is the limiting reactant.
To calculate the theoretical yield of HNO3 in grams, we multiply the moles of HNO3 produced by the molar mass of HNO3:
The molar mass of HNO3 is 63.01 g/mol.
Theoretical yield of HNO3 = Moles of HNO3 x Molar mass of HNO3
Theoretical yield of HNO3 = 43.45 mol x (63.01 g/mol) = 2,741.65 g
The theoretical yield of nitric acid is 2,741.65 grams.
To find the grams of excess reactant present after the reaction, we need to calculate the moles of excess reactant consumed and multiply it by its molar mass:
Moles of excess reactant = Moles of H2O - Moles of HNO3
Moles of excess reactant = 55.50 mol - 43.45 mol = 12.05 mol
Molar mass of H2O is 18.02 g/mol.
Grams of excess reactant = Moles of excess reactant x Molar mass of H2O
Grams of excess reactant = 12.05 mol x (18.02 g/mol) = 217.41 g
Therefore, there are 217.41 grams of excess reactant present after the reaction is complete.