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3 years ago

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Chemistry

1 answer:

4vir4ik [10]3 years ago

4 0

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In the reaction, Zn(s) + 2 HCl(aq) --> ZnCl2 (aq)+ H2(g), 25 grams of Zn are reacted with 17.5 g of HCl. How many grams of H2

Gennadij [26K]

Answer:

0.480 g of H₂ are produced, in the reaction.

Explanation:

This is the reaction:

Zn(s) + 2HCl → ZnCl₂ (aq) + H₂ (g)

We havethe mass of both reactants, so we must work with them to find out the limiting reactant and then, determine the amount of H₂ produced.

Let's convert the mass to moles ( mass / molar mass)

25 g / 65.41 g/mol = 0.382 moles Zn

17.5 g / 36.45 g/mol = 0.480 moles HCl

Ratio is 1:2, so 1 mol of Zn react with the double of moles of HCl.

0.382 moles of Zn would need the double of moles to react, so (0.382 .2) = 0.764 moles of HCl. → We only have 0.480 moles, so the acid is the limiting.

Now let's determine the moles of H₂ formed.

Ratio is 2:1, so If i take account the moles I have, I will produce the half of moles of my product.

0.480 moles / 2 = 0.240 moles of H₂ are produced.

To find out the mass, we must multiply mol . molar mass

0.240 mol . 2g/mol = 0.480 g

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A voltaic cell is based on the reduction of ag (aq) to ag(s) and the oxidation of sn(s) to sn2 (aq).

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Half-reaction for the cell's anode is given below:

Anode : $\\Sn(s) \rightarrow Sn^{2+} (aq) +2e^{-}$

The anode is defined as the electrode at which electrons leave the cell and oxidation occurs, and the cathode as the electrode at which electrons enter the cell and reduction occurs. The anode is usually the positive side.

Learn more about anode here:

brainly.com/question/4052514

Your given question is quite incomplete here is complete question.

A voltaic cell is based on the reduction of _ Agt(aq) to Ag(s) and the oxidation of Sn(s) to Sn2+(aql) : Part 1 Include the phases of all species in the chemical equation: (aqh Anode: Sn(s) Sn?+ (aq)

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