A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and of sulfate ions in the final solution.

NH

+

4

:

0.272

M

SO

2

−

4

:

0.136

M

Explanation:

We can solve this problem using some molarity calculations:

molarity

=

mol solute

L soln

We should convert the given mass of

(NH

4

)

2

SO

4

to moles using its molar mass (calculated to be

132.14

g/mol

):

10.8

g (NH

4

)

2

SO

4

⎛

⎝

1

l

mol (NH

4

)

2

SO

4

132.14

g (NH

4

)

2

SO

4

⎞

⎠

=

0.0817

mol (NH

4

)

2

SO

4

This is the quantity present in

100

mL soln

, so let's calculate the molarity of the solution (converting volume to liters):

molarity

=

0.0817

l

mol (NH

4

)

2

SO

4

0.100

l

L soln

=

0.817

M

10

mL

of this solution is added to

50

mL H

2

O

, which makes a

60

-

mL

total solution.

We can now use the dilution equation

M

1

V

1

=

M

2

V

2

to find the molality of the new,

60

-

mL

solution:

(

0.0817

M

)

(

10

l

mL

)

=

(

M

2

)

(

60

l

mL

)

M

2

=

(

0.817

M

)

(

10

mL

)

60

mL

=

0.136

M

This means that there are

0.136

moles of

(NH

4

)

2

SO

4

per liter of solution.

Let's recognize that

1

mol (NH

4

)

2

SO

4

contains

2

mol NH

+

4

1

mol SO

2

−

4

The concentrations of each ion is thus

(

2

)

(

0.136

M

)

=

0.272

M

NH

+

4

(

1

)

(

0.136

M

)

=

0.136

M

SO

2

−

4

Explanation:

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: