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olganol [36]
3 years ago
10

How many grams of silver chloride are produced from 5g of Silver Nitrate reacting with an excess of barium chloride?

Chemistry
1 answer:
monitta3 years ago
3 0
4.22g
1. Work out the equation and balance
2AgNo3 +BaCl2 ---> 2AgCl + Ba(No3)2

2. Work out formula mass of silver nitrate (should = 169.9)

3. Calculate the number of moles by doing moles = mass / formula mass so 5 divided by 169.9 = 0.0294 moles.


4. Check ratio - here the ratio is 2:2 i.e. 2 moles of silver nitrate to 2 moles of silver chloride so the moles will be the same so the moles of silver chloride is also 0.0294

5. Work out the formula mass of AgCl (always ignore big numbers at the start when working out formula mass) = 143.4

6. Work out mass by doing equation moles = mass/formula mass so
mass = moles x formula mass
mass = 0.0294 x 143.4
mass = 4.22g (3sf or 2dp)
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Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
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The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

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3 years ago
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