Chemistry, asked by manvendrakhichi269, 2 months ago

17. 100 mL of 0.1 M NaCl solution is mixed with 100 mL of 0.2 M AgNO, solution. The
reaction is as follows:
NaCl(aq) + AgNO3(aq)
→ NaNO3(aq) + AgCl(s)
(a)
Identify the limiting reagent.
(b)
Find out the mass of AgCl precipitate formed. (Molar mass of AgCl = 143.5 g
mol-1)
1+1=2​

Answers

Answered by humaidafridi18
1

Answer:

In the reaction shown above, if we mixed 123 mL of a 1.00 M solution of NaCl with 72.5 mL of a 2.71 M solution of AgNO 3 , we could calculate the moles (and hence, the mass) of AgCl that will be formed as follows: To determine the concentration of potassium nitrate in the final solution, we need to note that two moles of potassium nitrate are formed for every mole of PbI 2 , or a stoichiometric ratio of \[\left ( \frac{2\: moles\: KNO_{3}}{1\: mole\: PbI_{2}} \right )\]

Explanation:

Answered by AntolinOtaku23
0

Answer:

17.6 g AgCl(s) is formed in the reaction.

Explanation:

First, we need to write out our balanced reaction equation:

A

g

N

O

3

(

a

q

)

+

N

a

C

l

(

a

q

)

A

g

C

l

(

s

)

+

N

a

N

O

3

(

a

q

)

The next step, as in any calculation involving stoichiometry, is to determine our limiting reactant. We can do this by converting both of our reactants into moles:

123 mL

N

a

C

l

×

1 L

1000 mL

×

1.00 mol

N

a

C

l

1 L

=

0.123 mol

N

a

C

l

72.5 mL

A

g

N

O

3

×

1 L

1000 mL

×

2.71 mol

A

g

N

O

3

1 L

=

0.196 mol

A

g

N

O

3

We can see from our reaction equation that AgNO3 and NaCl react in a 1:1 ratio. Because there are fewer moles of NaCl present in solution, NaCl is our limiting reactant. We can now solve for the mass of AgCl formed:

123 mL

N

a

C

l

×

1 L

1000 mL

×

1.00 mol

N

a

C

l

1 L

×

1 mol

A

g

C

l

1 mol

N

a

C

l

×

143 g

1 mol

A

g

C

l

=

17.6 g

A

g

C

l

,

Therefore, 17.6 g AgCl(s) is formed in the reaction.

To sum up: we converted to each reactant’s moles by using the given concentrations as conversion factors, expressing molarity as mol/L; once we found our limiting reactant, we converted through to grams of AgCl formed.

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