Question

Balance the following equation ___Ba3(PO4)2 + ___HNO3 ---> ___Ba(NO3)2 + ___ H3PO4

How many grams of each product can be made if 13.55g of Ba3(PO4)2 and 11.21 of HNO3 are reacted? (don't forget about the limiting reagent)

Answer 1

As HNO3 is provided more in mass that's why barium phosphate is limiting in nature and used for further calculations.

Answer 2

To balance the chemical equation, we need to make sure that the same number of atoms of each element are present on both sides of the equation. Here's the balanced equation:

3 Ba3(PO4)2 + 6 HNO3 ---> 2 Ba(NO3)2 + 3 H3PO4

Now, let's calculate the number of moles of each reactant:

1. Moles of Ba3(PO4)2: Given mass of Ba3(PO4)2 = 13.55 g Molar mass of Ba3(PO4)2 = 3 * (137.33 g/mol Ba) + 2 * (30.97 g/mol P) + 8 * (16.00 g/mol O) = 601.93 g/mol Moles of Ba3(PO4)2 = 13.55 g / 601.93 g/mol ≈ 0.0225 mol

2. Moles of HNO3: Given mass of HNO3 = 11.21 g Molar mass of HNO3 = 1 * (1.01 g/mol H) + 1 * (14.01 g/mol N) + 3 * (16.00 g/mol O) = 63.02 g/mol Moles of HNO3 = 11.21 g / 63.02 g/mol ≈ 0.1779 mol

Now, let's determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction, thus limiting the amount of product formed. To find the limiting reagent, we compare the mole ratio of the reactants to the stoichiometry of the balanced equation:

Mole ratio: Ba3(PO4)2 : HNO3 = 0.0225 mol : 0.1779 mol ≈ 1 : 7.91

From the balanced equation, we can see that the mole ratio should be 1 : 6. So, HNO3 is the limiting reagent since it would require 6 times fewer moles to react with all of the Ba3(PO4)2.

Now, let's calculate the moles of each product that can be formed:

1. Moles of Ba(NO3)2: From the balanced equation, the mole ratio between Ba3(PO4)2 and Ba(NO3)2 is 3 : 2. Moles of Ba(NO3)2 = 0.0225 mol Ba3(PO4)2 * (2 mol Ba(NO3)2 / 3 mol Ba3(PO4)2) ≈ 0.0150 mol

2. Moles of H3PO4: From the balanced equation, the mole ratio between HNO3 and H3PO4 is 6 : 3 (or 2 : 1). Moles of H3PO4 = 0.1779 mol HNO3 * (3 mol H3PO4 / 6 mol HNO3) ≈ 0.08895 mol

Finally, let's calculate the mass of each product:

1. Mass of Ba(NO3)2: Molar mass of Ba(NO3)2 = 1 * (137.33 g/mol Ba) + 2 * (14.01 g/mol N) + 6 * (16.00 g/mol O) = 261.34 g/mol Mass of Ba(NO3)2 = 0.0150 mol * 261.34 g/mol ≈ 3.92 g

2. Mass of H3PO4: Molar mass of H3PO4 = 3 * (1.01 g/mol H) + 1 * (30.97 g/mol P) + 4 * (16.00 g/mol O) = 98.00 g/mol Mass of H3PO4 = 0.08895 mol * 98.00 g/mol ≈ 8.71 g

Therefore, approximately 3.92 grams of Ba(NO3)2 and 8.71 grams of H3PO4 can be made from the given reactants when 13.55 g of Ba3(PO4)2 and 11.21 g of HNO3 are reacted, with HNO3 being the limiting reagent.