5)When solutions of silver nitrate and potassium chloride are mixed, silver chlo

Question

5)When solutions of silver nitrate and potassium chloride are mixed, silver chloride precipitates out of solution according to the equation

AgNO3(aq)+KCl(aq)AgCl(s)+KNO3(aq)

a)What mass of silver chloride can be produced from 1.17 L of a 0.122 M solution of silver nitrate?

b)The reaction described in Part A required 3.39 L of potassium chloride. What is the concentration of this potassium chloride solution?

5)Consider the precipitation reaction: Li2S(aq)+Co(NO3)2(aq)2LiNO3(aq)+CoS(s)

a)What volume of 0.160 M Li2S solution is required to completely react with 125 mL of 0.160 M Co(NO3)2?

6)What is the minimum amount of 6.8 M H2SO4 necessary to produce 27.3 g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)Al2(SO4)3(aq)+3H2(g)

7)Consider the precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)Cu3(PO4)2(s)+6NaCl(aq)

a) What volume of 0.180 M Na3PO4 solution is necessary to completely react with 87.1 mL of 0.122 M CuCl2?

Explanation / Answer

5)

AgNO3(aq) + KCl(aq)-------------------> AgCl(s)+KNO3(aq)

(a) moles of AgNO3 = molarity x volume = 0.122 x 1.17

                                   = 0.143

1 mole AgNO3 = 1 mole AgCl

0.143 mol AgNO3 = 0.143 mole AgCl

AgCl moles = AgCl mass / AgCl molar mass

AgCl mass = AgCl moles x AgCl molar mass

                 = 0.143 x 143.32

                 = 20.5 g

AgCl mass = 20.5 g

(b)

from balanced equation moles of AgNO3 = moles of KCl

0.122 x 1.77 = C x 3.39

C = 0.0637 M

concentration of KCl = 0.0637 M

problem:

5)Consider the precipitation reaction: Li2S(aq)+Co(NO3)2(aq)2LiNO3(aq)+CoS(s)

a)What volume of 0.160 M Li2S solution is required to completely react with 125 mL of 0.160 M Co(NO3)2?

M1 V1 = M2V2

0.160 x V1 = 0.160 x 125

V1 = 125 ml

volume of Li2S = 125 ml

6)What is the minimum amount of 6.8 M H2SO4 necessary to produce 27.3 g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)Al2(SO4)3(aq)+3H2(g)

moles of H2 = 27.3 / 2 = 13.65

     3 mole H2SO4 gives = 3 mole H2 gas

13.65 moles H2SO4 = 13.65 mole H2 gas

H2SO4 moles = 13.65

moles = molarity x volume

13.65 = 6.8 x volume

volume = 2 L

2 Litres of H2SO4 is needed

7)Consider the precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)Cu3(PO4)2(s)+6NaCl(aq)

a) What volume of 0.180 M Na3PO4 solution is necessary to completely react with 87.1 mL of 0.122 M CuCl2?

M1 V1 / n1 = M2 V2 /n2

0.180 x V1 / 2 = 0.122 x 87.1 / 3

V1 = 39.35 ml

volume of Na3PO4 = 39.35 ml

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