Based on the equation of the reaction and the data provided,
- NH4NO3 is the limiting reactant
- mass of Na3PO4 left is 29.375 g
- 18.75 g of (NH4)3PO4 is produced
- 31.875 g of NaNO3 is produced
<h3>What are limiting reactants?</h3>
A limiting reactant is a reactant which is used up in a reaction after which the reaction stops.
In the given reaction:
3 NH4NO3 + Na3PO4 → (NH4)3PO4 + 3 NaNO3
The limiting reactant is determined from the stoichiometry of the eqaution.
Moles of reactant = mass/molar mass
Molar mass of NH4NO3 = 80 g/mol
Molar mass of Na3PO4 = 165 g/mol
Molar mass of (NH4)3PO4 = 150 g/mol
Molar mass of NaNO3 = 85 g/mol
From the equation of the reaction, 240 g (3 × 80) of NH4NO3 is required to react with 165 g of Na3PO4
There are only 30.0 g of NH4NO3 reacting with 50.0 g of Na3PO4
30 g of Na3PO4 will react with 30 × 165/240 = 20.625 g of Na3PO4
Therefore, NH4NO3 is the limiting reactant
Na3PO4 is the excess reactant
mass of Na3PO4 left = 50 - 20.625
mass of excess reactant left = 29.375 g
moles of NH4NO3 in 30 g = 30/80 = 0.375 moles
3 moles of NH4NO3 produces 1 mole of (NH4)3PO4
0.375 moles of NH4NO3 will produce 0.375 × 1/3 = 0.125 moles of (NH4)3PO4
mass of 0.125 moles of (NH4)3PO4 = 0.125 × 150
mass of (NH4)3PO4 produced = 18.75 g of (NH4)3PO4
3 moles of NH4NO3 produces 3 moles of NaNO3
0.375 moles of NH4NO3 will produce 0.375 moles of NaNO3
mass of 0.375 moles of NaNO3 = 0.375 × 85
mass of NaNO3 produced = 31.875 g of NaNO3
Learn more about limiting reactant at: brainly.com/question/24945784
