1. Calculate the new molarity that results when 500 mL of a 0.5 M Ca(NO 3 ) 2 so

Question

1.      Calculate the new molarity that results when 500 mL of a 0.5 M Ca(NO3)2 solution is diluted with water to a final total volume of 1.7 L. SHOW WORK

2.      A 30.0 mL of a 0.20 M potassium bromide solution reacts with a 0.10 M Mercury (II) nitrate solution. SHOW WORK

a)      Write the balanced equation for this reaction.

b)      What is the volume of the mercury (II) nitrate solution?

C)      What is the mass of the precipitate produced?

3.      A 9.3 g sample of chromium (III) nitrate is placed in 30.0 mL of a 0.100 M barium hydroxide solution. SHOW WORK

a.       Write the balanced equation for this reaction.

b.      What is the molarity of the chromium (III) nitrate solution before the reaction occured?

c.       What is the mass of the precipitate produced?

Explanation / Answer

1

M1V1 = M2V2

M1 = 0.5 M
V1 = 0.5 L

M2 = ?
V2 = 1.7 L

(0.5)(0.5) = (1.7)M2

M2 = 0.147 M

2

a. 2 KBr(aq) + Hg(NO3)2(aq) --> HgBr2(s) + 2 KNO3(aq)

b. 0.0300 L X 0.20 mol/L = 0.0060 mol KBr X (1 mol Hg(NO3) / 2 mol KBr) = 0.0030 mol Hg(NO3)2

0.0030 mol Hg(NO3)2 / 0.10 mol/L = 0.030 L = 30.0 mL

c. 0.0030 mol Hg(NO3)2 ( 1 mol HgBr2/1 mol Hg(NO3)2) = 0.0030 mol HgBr2 X 360.4 g/mol = 10.8 g HgBr2

3

a). 2 Cr(NO3)3(aq) + 3 Ba(OH)2(aq) ------> 3 Ba(NO3)2(aq) + 2 Cr(OH)3(s)

b). mole of chromium (III) nitrate = 9.3 / 238 = 0.039 mole = 39 millimole
the molarity = 39 millimole / 30.0 milliliter = 1.3M

c). mole of barium hydroxide 30.0 mL * 0.100 M = 3 millimole
barium hydroxide is the limited reactant
mole of Cr(OH)3 = 2/3 * 3millimole = 2 millimole
mass of Cr(OH)3 = 2 millimole * 103gmol^-1 = 206 milligram = 0.206 g

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