Problem 3: Consider the following reaction: 3NH_(4)NO_(3)+Na_(3)PO_(4)arrow (NH_(4))_(3)PO_(4)+3NaNO_(3) Given: 50 grams of ammonium nitrate and 60 grams of sodium phosphate. What is the Maximum amount of (NH_(4))_(3)PO_(4) can be made? Your answer

Question

Answer 1

To determine the maximum amount of

that can be made, we need to identify the limiting reactant in the given reaction.The balanced chemical equation for the reaction is:

Given:- 50 grams of ammonium nitrate (

)- 60 grams of sodium phosphate (

)Step 1: Calculate the molar masses of the reactants and products.- Molar mass of

= 80.04 g/mol- Molar mass of

= 163.94 g/mol- Molar mass of

= 149.09 g/molStep 2: Calculate the number of moles of each reactant.- Moles of

= 50 g / 80.04 g/mol = 0.625 mol- Moles of

= 60 g / 163.94 g/mol = 0.366 molStep 3: Determine the limiting reactant.- According to the balanced equation, 3 moles of

react with 1 mole of

.- The ratio of moles of

to

is 0.625 / 0.366 = 1.71, which is greater than the required ratio of 3:1.- Therefore,

is the limiting reactant.Step 4: Calculate the maximum amount of

that can be made.- According to the balanced equation, 1 mole of

1 mole of

.- Since

is the limiting reactant, the maximum amount of

that can be made is equal to the number of moles of

, which is 0.366 mol.- The mass of

produced = 0.366 mol × 149.09 g/mol = 54.6 gTherefore, the maximum amount of

that can be made is 54.6 grams.

Question