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Fynjy0 [20]
3 years ago
13

How many moles of H 2 can be formed if a 3.24 g sample of M g reacts with excess H C l ?

Chemistry
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

0.135 mole of H2.

Explanation:

We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:

Mass of Mg = 3.24 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 3.24/24

Mole of Mg = 0.135 mole

Next, we shall write the balanced equation for the reaction. This is illustrated below:

Mg + 2HCl —> MgCl2 + H2

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.

Thus, 0.135 mole of H2 can be obtained from the reaction.

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Molar mass (NH4)2SO4
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Answer:

<u>132.15</u>

Explanation:

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Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
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Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

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3 years ago
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