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Softa [21]
2 years ago
14

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2AlCl3 ( aq ) +

3 H 2 ( g ) What volume of H 2 ( g ) is produced when 8.60 g Al ( s ) reacts at STP
Chemistry
1 answer:
Rufina [12.5K]2 years ago
6 0

Answer:

The answer to your question is  V = 19.9 L

Explanation:

Data

mass of Al = 8.60 g

volume of H₂ = ?

Balanced chemical reaction

                 2Al  +  6HCl  ⇒   2AlCl₃  +  3H₂

1.- Use proportions to calculate the mass of H₂ produced

Atomic mass of Al = 2 x 27 = 54 g

Atomic mass of H₂ = 6 x 1 = 6 g

                54 g of Al -------------------- 6 g of H₂

                  8 g  of Al -------------------  x

                  x = (8 x 6) / 54

                  x = 48 / 54

                  x = 0.89 g of H₂

2.- Calculate the volume of H₂

- Convert the H₂ to moles

                    1 g of H₂ --------------- 1 mol

                   0.89 g    ---------------- x

                    x = 0.89 moles

                    1 mol of H₂ -------------- 22.4 l

                   0.89 moles  ------------- x

                    x = (0.89 x 22.4) / 1

                   x = 19.9 L                        

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That would be false.
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Why is acetic anhydride added slowly to a dichloromethane solution of the other reactants?
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The acetic anhydride should be added slowly through a separatory funnel to other reactants so as to avoid a runaway reaction, also known as thermal runaway. By doing this the reaction mixture boils as gently as possible while stirring. This phenomenon occurs in situations where an increase in temperature changes the conditions in a way that leads to destructive result brought about by a further increase in temperature.
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3 years ago
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What is the E°cell for the cell represented by the combination of the following half-reactions? ClO4–(aq) + 8H+(aq) + 8e– Cl–(aq
Vinvika [58]

Answer:

The E°cell for the cell represented by the combination of the given half-reactions is 0.398 V

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

This is the type of reaction that occurs in this case.

ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

VO₂⁺(aq) + 2 H⁺(aq) + e⁻ ⇔ VO⁺(aq) + H₂O(l) E° = 0.991 V

In this case both are written as reductions, and their E ° values ​​as well. The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the first half-reaction expressed.  Therefore, to obtain a reaction, the second semi-reaction must be reversed to be an oxidation, maintaining its constant value. Then:

Reduction: ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

Oxidation: VO⁺(aq) + H₂O(l) ⇔ VO₂⁺(aq) + 2 H⁺(aq) + e⁻ E° = 0.991 V

<em>E°cell=Ereduction - Eoxidation</em>

E°cell=1.389 V - 0.991 V

<em>E°cell=0.398 V</em>

Then <u><em>the E°cell for the cell represented by the combination of the given half-reactions is 0.398 V.</em></u>

Another way of thinking is that, by inverting the second semi-reaction to be an oxidation, the value of E ° is reversed in the sign, unlike the previous case in which it was constant. Then:

Reduction: ClO₄⁻(aq) + 8 H⁺ (aq) + 8 e⁻ ⇔ Cl⁻ (aq) + 4 H₂O(l) E° = 1.389 V

Oxidation: VO⁺(aq) + H₂O(l) ⇔ VO₂⁺(aq) + 2 H⁺(aq) + e⁻ E° = -0.991 V

In this case:

E°cell=Ereduction + Eoxidation=

E°cell=1.389 V + (-0.991 V)=1.389 V-0.991 V

<em>E°cell=0.398 V</em>

Note that the result obtained is the same. This indicates that either of the two ways proposed is correct, and you will use the one that is most comfortable for you.

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Answer:

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According to the given reaction, 2 moles of As react with 6 moles of NaOH. Use this ratio to find how many moles of NaOH react with 492 moles of As.

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It means that you need 1476 moles of NaOH to use 492 moles of As all up.

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