Ammonia, NH3, reacts with Phosphoric acid to produce Ammonium hydrogenphosphate, (NH4)2HPO4.
What mass (in grams) of ammonia will react with 2.800E3 kg of Phosphoric acid?
To determine the mass of ammonia that will react with a given quantity of phosphoric acid, we need to start by balancing the equation for the reaction between ammonia (NH3) and phosphoric acid (H3PO4).
The balanced equation for this reaction is as follows:
3NH3 + H3PO4 -> (NH4)2HPO4
This means that for every 3 moles of ammonia, we need 1 mole of phosphoric acid to produce 1 mole of ammonium hydrogenphosphate.
Now, let's start by converting the given mass of phosphoric acid from kilograms to grams:
2.800E3 kg = 2.800E3 x 1000 g = 2.800E6 g
Next, we need to calculate the molar mass of phosphoric acid (H3PO4) and ammonium hydrogenphosphate ((NH4)2HPO4) to determine the moles of each.
Molar mass of H3PO4:
H = 1.01 g/mol
P = 30.97 g/mol
O = 16.00 g/mol
Molar mass of H3PO4 = (3 x 1.01 g/mol) + 30.97 g/mol + (4 x 16.00 g/mol) = 98.00 g/mol
Molar mass of (NH4)2HPO4:
N = 14.01 g/mol
H = 1.01 g/mol
P = 30.97 g/mol
O = 16.00 g/mol
Molar mass of (NH4)2HPO4 = (2 x (2 x 1.01 g/mol + 14.01 g/mol)) + 30.97 g/mol + (4 x 16.00 g/mol) = 132.14 g/mol
Next, we convert the mass of phosphoric acid to moles using its molar mass:
moles of H3PO4 = mass (g) / molar mass (g/mol)
moles of H3PO4 = 2.800E6 g / 98.00 g/mol = 2.857E4 mol
Using the balanced equation, we can see that for every 1 mole of phosphoric acid, we need 3 moles of ammonia.
Therefore, the moles of ammonia needed would be:
moles of NH3 = 3 x moles of H3PO4 = 3 x 2.857E4 mol = 8.571E4 mol
Finally, we can calculate the mass of ammonia in grams using its molar mass:
mass of NH3 = moles of NH3 x molar mass of NH3
mass of NH3 = 8.571E4 mol x 17.03 g/mol = 1.460E6 g
Therefore, the mass of ammonia that will react with 2.800E3 kg of phosphoric acid is 1.460E6 grams.
To determine the mass of ammonia that will react with 2.800E3 kg of Phosphoric acid, we need to use the balanced equation and the molar masses of ammonia and Phosphoric acid.
First, we need to write the balanced equation for the reaction between ammonia and Phosphoric acid:
NH3 + H3PO4 → (NH4)2HPO4
According to the balanced equation, the molar ratio between ammonia and Phosphoric acid is 1:1. This means that for every 1 mole of Phosphoric acid, we need 1 mole of ammonia.
Now, we can calculate the amount of Phosphoric acid in moles:
Molar mass of Phosphoric acid (H3PO4) = (1*1.0079) + (3*15.9994) + (1*30.9738) = 97.995 grams/mol
Mass of Phosphoric acid = 2.800E3 kg = 2.800E3 * 1000 g = 2.800E6 grams
Number of moles of Phosphoric acid = Mass of Phosphoric acid / Molar mass of Phosphoric acid
= 2.800E6 grams / 97.995 grams/mol
≈ 2.857E4 moles
Since the molar ratio between ammonia and Phosphoric acid is 1:1, the number of moles of ammonia required will be the same as the number of moles of Phosphoric acid.
Therefore, the mass of ammonia required can be calculated using the molar mass of ammonia (17.031 grams/mol):
Mass of ammonia required = Number of moles of ammonia * Molar mass of ammonia
= 2.857E4 moles * 17.031 grams/mol
≈ 4.864E5 grams
Therefore, approximately 4.864E5 grams (486,400 grams) of ammonia will react with 2.800E3 kg of Phosphoric acid.
2NH3 + H3PO4 ==> (NH4)2HPO4
mols H3PO4 ==> grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols H3PO4 to mols NH3.
Now convert mols NH3 to grams. g = mols x molar mass