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Chapter 4: Problem 22

What volume of \(0.4163 M\) barium chloride will react completely with (a) \(12.45 \mathrm{~mL}\) of \(1.732 \mathrm{M}\) sulfuric acid? (b) \(15.00 \mathrm{~g}\) of ammonium phosphate? (c) \(35.15 \mathrm{~mL}\) of \(1.28 \mathrm{M}\) potassium carbonate?

Short Answer

Expert verified
Question: Calculate the volume of 0.4163 M barium chloride required to react completely with each of the following: (a) 12.45 mL of 1.732 M sulfuric acid. (b) 15.00 g of ammonium phosphate. (c) 35.15 mL of 1.28 M potassium carbonate. Answer: (a) 51.9 mL of 0.4163 M barium chloride will react completely with 12.45 mL of 1.732 M sulfuric acid. (b) 363 mL of 0.4163 M barium chloride will react completely with 15.00 g of ammonium phosphate. (c) 108 mL of 0.4163 M barium chloride will react completely with 35.15 mL of 1.28 M potassium carbonate.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction of barium chloride with sulfuric acid is: $$\mathrm{BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl}$$
02

Determine the molar amount of sulfuric acid

We are given the volume (12.45 mL) and molarity (1.732 M) of sulfuric acid. Use these values to calculate the moles of sulfuric acid: $$\mathrm{moles\:of\:H_2SO_4 = 1.732\,M \times \frac{12.45\,mL}{1000\,mL/mol} = 0.0216\,mol}$$
03

Determine the molar amount of barium chloride

From the balanced chemical equation, we see that 1 mole of barium chloride reacts with 1 mole of sulfuric acid. So, we need 0.0216 moles of barium chloride to react with the given sulfuric acid.
04

Calculate the volume of barium chloride

We are given the molarity of barium chloride (0.4163 M). Use this value and the moles of barium chloride to find the volume: $$\mathrm{Volume\:of\:BaCl_2 = \frac{0.0216\,mol}{0.4163\,mol/L} = 0.0519\,L\: or\: 51.9\:mL}$$ So, 51.9 mL of 0.4163 M barium chloride will react completely with 12.45 mL of 1.732 M sulfuric acid. #Part (b) - 15.00 g of ammonium phosphate#
05

Write the balanced chemical equation

The balanced chemical equation for the reaction of barium chloride with ammonium phosphate is: $$\mathrm{3BaCl_2 + 2(NH_4)_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6NH_4Cl}$$
06

Determine the molar amount of ammonium phosphate

We are given the mass (15.00 g) of ammonium phosphate. First, we need to calculate its molar mass: $$\mathrm{Molar\:mass\:of\:(NH_4)_3PO_4 = 3(14.0+4.0)+1(31.0)+4(16.0) = 149\: g/mol}$$ Now, calculate the moles of ammonium phosphate: $$\mathrm{moles\:of\:(NH_4)_3PO_4 = \frac{15.00\, g}{149\, g/mol} = 0.1007\, mol}$$
07

Determine the molar amount of barium chloride

From the balanced chemical equation, we see that 3 moles of barium chloride reacts with 2 moles of ammonium phosphate. So, we need: $$\mathrm{moles\:of\:BaCl_2 = \frac{0.1007\, mol\: (NH_4)_3PO_4}{2} \times 3 = 0.1511\, mol}$$
08

Calculate the volume of barium chloride

We are given the molarity of barium chloride (0.4163 M). Use this value and the moles of barium chloride to find the volume: $$\mathrm{Volume\:of\:BaCl_2 = \frac{0.1511\, mol}{0.4163\, mol/L} = 0.363\, L\: or\: 363\: mL}$$ So, 363 mL of 0.4163 M barium chloride will react completely with 15.00 g of ammonium phosphate. #Part (c) - 35.15 mL of 1.28 M potassium carbonate#
09

Write the balanced chemical equation

The balanced chemical equation for the reaction of barium chloride with potassium carbonate is: $$\mathrm{BaCl_2 + K_2CO_3 \rightarrow BaCO_3 + 2KCl}$$
10

Determine the molar amount of potassium carbonate

We are given the volume (35.15 mL) and molarity (1.28 M) of potassium carbonate. Calculate the moles of potassium carbonate: $$\mathrm{moles\:of\:K_2CO_3 = 1.28\,M \times \frac{35.15\,mL}{1000\,mL/mol} = 0.0450\,mol}$$
11

Determine the molar amount of barium chloride

From the balanced chemical equation, we see that 1 mole of barium chloride reacts with 1 mole of potassium carbonate. So, we need 0.0450 moles of barium chloride to react with the given potassium carbonate.
12

Calculate the volume of barium chloride

We are given the molarity of barium chloride (0.4163 M). Use this value and the moles of barium chloride to find the volume: $$\mathrm{Volume\:of\:BaCl_2 = \frac{0.0450\, mol}{0.4163\, mol/L} = 0.108\, L\: or\: 108\: mL}$$ So, 108 mL of 0.4163 M barium chloride will react completely with 35.15 mL of 1.28 M potassium carbonate.

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