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3 years ago

What is the percent of oxygen by mass in a pure sample of fecr2o4? 1. 52% 2. 13% 3. 37% 4. 7.1% 5. 29%?

1 answer:

6 0

Let us suppose that we have 1 mol of FeCr2O4. I'm going to use approximate masses because I have no idea what your periodic table will say. Just put in the exact masses from your periodic table.

Fe = 56
Cr*2 = 2*52 = 104
O4 = 4*16 = 64
===========
Total = 56 + 104 + 64 = 224
The % oxygen = (64 / 224) * 100 = 28.5 % but none of your answers match this. Perhaps you are talking about Fe2(Cr2O4)3 The brackets make all the difference in the world.

Without going through all the detail I did before, The molecular mass is
Fe * 2 = 112
Cr * 6 = 312
O * 12 = 192
The total molecular mass = 616

The % Oxygen = (192 / 616) * 100 = 31% roughly. That answer isn't there either.

Let's wait and see who else answers.

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In the equation below

yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

$\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}$

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)

The ratio between NH₃ and Cu is 2:3.
The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

The amount of N₂ produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}$

The amount of Cu produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}$

And the amount of H₂O produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}$

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

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Answer:

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alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

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jolli1 [7]

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