Answer to Question #183487 in General Chemistry for Krissy

Question #183487

Solution Stoichiometry Problem Set

3.   25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper (II) sulfate.  How many grams of  copper (II) hydroxide will precipitate?  (7 marks)

4. What volume of 0.415 M silver nitrate will be required to precipitate as silver bromide all the bromide ion in 35.0 mL of 0.128 M calcium bromide? (4 marks) due 4/19/2021 12am

Expert's answer

1. CuSO4 + 2NaOH––> Na2SO4 + Cu(OH)2

Moles of NaOH = CxV= 0.350 x 25.0/1000 = 8.75x10-3mol

Moles of CuSO4= CxV= 0.125 x 45.0/1000 = 5.625x10-3mol

Molar mass of NaOH= 40g/mol

Molar mass of CuSO4= 159.5g/mol

Molar mass of Cu(OH)2= 97.5g/mol

Mass of NaOH= mole x molar mass= 8.75x10-3 x 40= 0.35g

Mass of CuSO4= 5.625x10-3 x 159.5= 0.897g

Let's find the limiting reagent that determines the amount of Cu(OH)2 formed. From the balanced equation

159.5g of CuSO4 reacts with 80g of NaOH

0.897g of CuSO4 should react with 80/159.5 x 0.897= 0.45g

But only 0.35g of NaOH is available, therefore it is the limiting reagent.

80g of NaOH gives 97.5g of Cu(OH)2

0.35g of NaOH will give 97.5/80 x 0.35= 0.43g of Cu(OH)2

2. CaBr2 + 2AgNO3---> 2AgBr + Ca(NO3)2

Mole of CaBr2= 0.128 x 35.0/1000= 4.48 x 10-3mol

From the balanced equation,

1mol of CaBr2 reacts with 2mol of AgNO3

4.48x10-3 mol of CaBr2 will react with 2/1 x 4.48 x 10-3= 8.96x10-3mol of AgNO3

Volume of AgNO3= mole/concentration = 8.96x10-3/0.415 = 0.0216L = 21.6ml of AgNO3

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