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3 years ago

14

I Solid iron(III) oxide is produced when hot iron filings are exposed to pure oxygen gas.

1 answer:

Sergeu [11.5K]3 years ago

8 0

The unbalanced equation is:
Fe(s) + O2(g) —> Fe2O3(s)

To balance oxygens, multiply O2 by 3 and Fe2O3 by 2.

Fe(s) + 3O2(g) —> 2Fe2O3(s)

Balance your iron by multiplying Fe by 4, since you have 4 Fe in the products now.

4Fe(s) + 3O2(g) —> 2Fe2O3(s)

You might be interested in

A student is given 1.525 g of pure CuO. To recover the Cu present in the compound, the dark powdery solid was dissolved in 15.0

ikadub [295]

Answer:

The amount of Mg was enough

Explanation:

In this case, we have to start with the <u>reaction</u> between $Mg$ and $CuO$ , so:

$CuO~+~Mg~->~MgO~+~Cu$

If we check <u>the reaction is already balanced</u>. Now, we can do some stoichiometry to calculate the amount of Mg. The first step is the number of moles of $CuO$ . To this we have to calculate the molar mass of $CuO$ first, so:

Cu: 63.55 g/mol and O: 16 g/mol. So, (63.55+16)= 79.55 g/mol.

Now, we can calculate the moles:

$1.525~g~CuO\frac{1~mol~CuO}{79.55~g~CuO}=0.0192~mol~CuO$

The <u>molar ratio</u> between $Mg$ and $CuO$ is 1:1, so:

$0.0192~mol~CuO=0.0192~mol~Mg$ .

Now we can <u>calculate the mass of M</u>g if we know the atomic mass of Mg (24.305 g/mol). So:

$0.0192~mol~Mg\frac{24.305~g~Mg}{1~mol~Mg}=0.466~g~Mg$

<u>With this in mind, the student added enough Mg to recover all the Cu.</u>

Note: The HCl doesn't take a role in the reaction. The function of HCl is to dissolve the $CuO$ .

I hope it helps!

3 0

3 years ago

For the results of an experiment to be truly reliable, _____

Igoryamba

Answer:

bobux fduyfgyusgfuyogdsfygdfy

Explanation:

5 0

3 years ago

Read 2 more answers

What mass of H2 is needed to react with 8.5 g of O2 according to the following equation G plus O2 (G)plus H2 ( G)?

Rzqust [24]

The correct answer is number 1

5 0

3 years ago

Tin has ten stable isotopes. The heaviest, 124Sn, makes up 5.80% of naturally occuring tin atoms. How many atoms of 124Sn are pr

matrenka [14]

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = $6.023X10^{23}X0.691=4.16X10^{23}$

the percentage of heaviest Sn is 5.80%

So the total atoms of $Sn^{124}$ = 5.80% X $4.16X10^{23}$

Total atoms of $Sn^{124}$ = $2.41X10^{22}$ atoms

the mass of $Sn^{124}$ will be = $\frac{2.41X10^{22}X124}{6.023X10^{23}}=4.96g$

5 0

4 years ago

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.

bagirrra123 [75]

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>

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I hope it helps!

5 0

3 years ago