Question

NH4NO3 + Na3PO4 → (NH4)3PO4 + NaNO3

Answer 1

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

What are limiting reactants?

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

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