   Chapter 15, Problem 140CP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
87 views

# A 500.0-mL sample of 0.200 M sodium phosphate is mixed with 400.0 mL of 0.289 M barium chloride. What is the mass of the solid produced’?

Interpretation Introduction

Interpretation:

The mass of solid produced on mixing given solutions of sodium phosphate and barium chloride is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given as:

M=nV

Where,

• n represents the number of moles of the solute.
• V represents the volume of the solution.

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting reagents limits the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Explanation

The volume of 0.200M sodium phosphate is 500.0mL.

The volume of 0.289M barium chloride is 400.0mL.

The molarity of a solution is given as:

M=nV

Where,

• n represents the number of moles of the solute.
• V represents the volume of the solution.

Rearrange the above equation for the value of n.

n=MV    (1)

Substitute the value of molarity and volume of Na3PO4 solution in equation (1).

n=0.200M1molL11M500.0mL1L1000mL=0.100mol

The number of moles of Na3PO4 present in the reaction mixture is 0.100mol.

Substitute the value of molarity and volume of BaCl2 solution in equation (1).

n=0.289M1molL11M400.0mL1L1000mL=0.1156mol

The number of moles of BaCl2 present in the reaction mixture is 0.1156mol.

The balanced reaction between sodium phosphate and barium chloride is represented as:

2Na3PO4aq+3BaCl2aqBa3PO42s+6NaClaq

Two moles of Na3PO4 reacts with three moles of BaCl2 to produce one mole of Ba3PO42 and six moles of NaCl. Therefore, the relation between the number of moles of Na3PO4 and BaCl2 reacts is given as:

nBaCl2=3nNa3PO42

Where,

• nBaCl2 represents the number of moles of BaCl2.
• nNa3PO4 represents the number of moles of Na3PO4.

Substitute the value of number of moles of Na3PO4 in the above equation

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