galanis513
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NH4NO3 + Na3PO4 → (NH4)3PO4 + NaNO3
Which reactant is limiting, assuming we started with 30.0 grams of ammonium
nitrate and 50.0 grams of sodium phosphate. What is the mass of each product
that can be formed? What mass of the excess reactant(s) is left over?

Sagot :

mickymike92

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

Learn more about limiting reactant at: https://brainly.com/question/24945784

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