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Section 9.2 Solution Stoichiometry
Solutions for Practice Problems
Student Edition page 417
11. Practice Problem (page 417) If 8.5 g of pure ammonium phosphate, (NH 4 ) 3 PO 4 (s), is dissolved in distilled water to make 400 mL of solution, what are the concentrations (in moles per litre) of the ions in the solution?
What Is Required? You need to find the molar concentration, c , of the ions in a solution of ammonium phosphate.
What Is Given? You know the volume of the ammonium phosphate solution: 400 mL You know the mass of ammonium phosphate, (NH 4 ) 3 PO 4 (s): 8.5 g
Plan Your Strategy Write the balanced chemical equation for the dissolution of ammonium phosphate, (NH 4 ) 3 PO 4 (s). Determine the molar mass of (NH 4 ) 3 PO 4 (s).
Calculate the amount in moles of (NH 4 ) 3 PO 4 (s) using the relationship.
n n M Convert the volume from millilitres to litres: 1 mL = 1 × 10–3^ L Calculate the concentration of (NH 4 ) 3 PO 4 (aq) using the relationship (^). n c V Equate the mole ratios and cross multiply to solve for n , the amount in moles of (NH 4 ) 3 PO 4 (s).
Act on Your Strategy Balanced equation: (NH 4 ) 3 PO 4 (s) ĺ 3NH 4 +(aq) + PO 4 3–(aq) Mole ratio: 1 mole 3 moles 1 mole
Molar mass, M , of (NH 4 ) 3 PO 4 (s): NH 4 3 PO 4 3 N^12 H^1 P^4 O 3 14.01g/mol 12 1.01 g/mol 1 30.97 g / mol 4 16.00 g/mol 149.12 g/mol
M M M M M
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Amount in moles, n , of (NH 4 ) 3 PO 4 (s):
NH 4 3 PO 4
8.5 g
m n M
149.12 g 5.700 1 –
/mo 0 o
l u m l
Amount in moles, n , of NH 4 +(aq):
4
4
4 3 4 4 3 4 (^4) NH
4 4 NH
1 mol (NH ) PO 0.005700 mol (NH ) PO 3 mol NH
3 mol NH 0.005700 mol (NH ) PO
n
n
(^) u 3 4 1 mol (NH ) PO 4 3 4 0.0171 mol
Amount in moles, n , of PO 4 3–(aq):
4 3–
4 3–
4 3 4 4 3 4 3– (^4) PO
4 4 P
3 O
1 mol (NH ) PO 0.005700 mol (NH ) PO 1 mol PO
1 mol PO 0.005700 mol (NH ) PO
n
n
(^) u 3 4 1 mol (NH ) PO 4 3 4 0.005700 mol
Volume of solution: V 400 mL × 1 × 10 –3L/ mL 0.400 L
Concentration of NH 4 +( aq):
1.71 10 –2 mol
0.04275 mol/L 0.04 mol/L
L
n c V u
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Amount in moles, n , of the precipitate, FeS(s): FeS 3 2 3 2 3 2 FeS
1 mol FeS 1 mol Fe(NO ) 0.0125 mol Fe(NO ) 1 mol FeS 0.0125 mol Fe(NO )
n
n
u 1 mol Fe(NO ) 3 2 0.0125 mol
Molar mass, M , of the precipitate, FeS(s): FeS 1 Fe 1 S 1(55.85 g/mol) 1(32.07 g/mol) 87.92 g/mol
M M M
Mass, m , of FeS(s): FeS 0.0125 mol
m n u M u 87.92 g/ mol 1.099 g 1.10 g
The precipitate is iron(II) sulfide, FeS(s), and the maximum mass that can be collected from the reaction is 1.10 g.
Check Your Solution The units for amount and concentration are correct. The answer has three significant digits and seems reasonable.
15. Practice Problem (page 417) What mass of silver chloride, AgCl(s), can be precipitated from 75 mL of 0. mol/L silver nitrate, AgNO 3 (aq), by adding excess magnesium chloride, MgCl 2 (aq)?
What Is Required? You need to calculate the mass of silver chloride that will precipitate in a reaction.
What Is Given? You know the volume of the silver nitrate solution: 75 mL You know the initial concentration of silver nitrate: 0.25 mol/L You know the other reactant is aqueous magnesium chloride, MgCl 2 (aq).
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Plan Your Strategy Predict the other product that forms in this double displacement reaction. Write the balanced equation for the reaction. Convert the volume to litres: 1 mL = 1 × 10–3^ L Calculate the amount in moles of silver nitrate using the relationship n c u V. Equate the mole ratios and solve for the amount in moles of precipitate. Use the periodic table to determine the molar mass of the precipitate. Calculate the mass of precipitate using the relationship m n u M.
Act on Your Strategy The other product is magnesium nitrate, Mg(NO 3 ) 2 (aq).
Balanced equation: MgCl 2 (aq) + 2AgNO 3 (aq) ĺ Mg(NO 3 ) 2 (aq) + 2AgCl(s) Mole ratio: 1 mole 2 moles 1 mole 2 moles
Volume of solution: V 75 mLu 1 u 10 –3L/ mL 0.075 L
Amount in moles, n , of AgNO 3 (aq): AgNO 3 0.25 mol / L
n c u V
u 0.075 L 0.01875 mol
Amount in moles, n , of precipitate, AgCl(s):
3 3 3
AgCl
AgCl
2 mol AgCl 2 mol AgNO 0.01875 mol AgNO 2 mol AgCl 0.01875 mol AgNO
n
n
u 2 mol AgNO 3 0.01875 mol
Molar mass, M , of the precipitate, AgCl(s): AgCl 1 Ag 1 Cl 1 107.87 g/mol 1(35.45 g/mol) 143.32 g/mol
M M M
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Mass, m , of NaCl(s): m NaCl (^) 0.213508 molu58.44 g/ mol 12.477 g 12.5 g
The mass of sodium chloride in the 1.0 L volumetric flask is 12.5 g.
Check Your Solution The units for amount and concentration are correct. The answer has three significant digits and seems reasonable.
20. Practice Problem (page 417) Food manufacturers sometimes add calcium acetate, Ca(CH 3 COO) 2 (s), to sauces as a thickening agent. When analyzed, a 250 mL solution of calcium acetate was found to contain 0.200 mol of acetate ions. a. Find the molar concentration of the calcium acetate solution. b. What mass of calcium acetate was dissolved to make the solution?
a. molar concentration What Is Required? You need to find the molar concentration of a calcium acetate solution.
What Is Given? You know the chemical formula for calcium acetate: Ca(CH 3 COO) 2 (s) You know the amount in moles of acetate ions, CH 3 COO–(aq): 0.200 mol You know the volume of the solution: 250 mL
Plan Your Strategy Use the mole ratio of acetate ions to calcium acetate to determine the amount in moles of calcium acetate. Convert the volume from millilitres to litres: 1 mL = 1 × 10–3^ L
Calculate the concentration of calcium acetate using the relationship. n c V
Act on Your Strategy The chemical formula for calcium acetate, Ca(CH 3 COO) 2 (s), indicates that there are two CH 3 COO–^ ions for one formula unit of Ca(CH 3 COO) 2 (s).
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Amount in moles, n , of Ca(CH 3 COO) 2 (s):
3 2
3 2
3 2 Ca (CH COO)
3 2 3 Ca (CH COO)
1 mol Ca(CH COO) 2 mol CH COO 0.200 mol CH COO
1 mol Ca(CH COO) 0.200 mol CH COO
n
n
u
2 mol CH COO 3 0.100 mol
Volume of solution: V 250 mLu 1 u 10 –3L/ mL 0.250 L
Concentration of Ca(CH 3 COO) 2 (aq):
0.100 mol 0.250 L 0.40 mol/L
n c V
The concentration of calcium acetate is 0.40 mol/L.
b. mass of calcium acetate What Is Required? You need to find the mass of calcium acetate in 250 mL of solution.
What Is Given? You know the molar concentration: 0.40 mol/L You know the volume of solution: 0.250 L
Plan Your Strategy Calculate the amount in moles of calcium acetate using the relationship the relationship n c u V. Use the periodic table to find the molar mass of Ca(CH 3 COO) 2 (s). Calculate the mass of calcium acetate using the relationship m n u M.
