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4 years ago

C) If 0.66 mole of iron (III) oxide were produced from the reaction, that must mean that how many mole

Chemistry

2 answers:

Fudgin [204]4 years ago

4 0

Answer:

If there is 0.66 moles of iron(III)oxide produced, there reacte 0.99 moles of oxygen (O2)

Explanation:

Step 1: Data given

Number of moles iron (III) oxide (Fe2O3) = 0.66 moles

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate moles of oxygen (O2)

For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3

For 0.66 moles Fe2O3 produced, we need 3/2 * 0.66 = 0.99 moles of O2

If there is 0.66 moles of iron(III)oxide produced, there reacte 0.99 moles of oxygen (O2)

Margarita [4]4 years ago

4 0

Answer:

0.99 moles of O₂

Explanation:

Let's consider the reaction between iron and oxygen to produce iron (III) oxide.

4 Fe + 3 O₂ → 2 Fe₂O₃

The molar ratio of oxygen to iron (III) oxide is 3:2. The moles of oxygen that reacted to form 0.66 moles of iron (III) oxide are:

0.66 mol Fe₂O₃ × (3 mol O₂/2 mol Fe₂O₃) = 0.99 mol O₂

0.99 moles of O₂ reacted to produce 0.66 moles of Fe₂O₃.

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A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac

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Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

$pH = pKa + log(\frac{[KF]}{[HF]})$

$pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04$

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

$\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol$

$\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles$

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

$\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles$

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)

The number of moles of HF after the reaction of KF with HCl is:

$\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles$

And the concentration of HF after the reaction of KF with HCl is is:

$C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L$

Now, from the equilibrium of equation (1) we have:

$Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}$

$Ka = \frac{x^{2}}{0.531 - x}$ (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

$pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73$

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

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