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Ilia_Sergeevich [38]
3 years ago
11

Assume 3 molecules of Fe react according to the following equation. 3Fe + 4H2O Fe3O4 + 4H2 How many molecules of H2 are produced

? How many water molecules are required? What is the mole ratio of Fe to H2O? How many hydrogen molecules (H2) are involved in this reaction?
Chemistry
2 answers:
user100 [1]3 years ago
6 0

How many molecules of H2 are produced?: 4 molecules

How many water molecules are required?: 4 molecules

What is the mole ratio of Fe to H2O?: 3/4

How many hydrogen molecules (H2) are involved in this reaction?: 4 molecules

Wewaii [24]3 years ago
5 0
Chemical reaction:  3Fe + 4H₂O → Fe₃O₄ + 4H₂<span>.
From chemical reaction:
1) n(Fe) : n(H</span>₂) = 3 : 4.
Four molecules of hydrogen are produced.
2) n(Fe) : n(H₂O) = 3 : 4.
Four molecules of water are required, because the mole ratio of Fe to H₂O is 3:4.
3) Hydrogen molecules are not<span> involved in this reaction, but produced (4 molecules).</span>
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A

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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
saul85 [17]

Answer :

AgBr should precipitate first.

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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2 years ago
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