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White raven [17]
2 years ago
10

A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which

Chemistry
2 answers:
lilavasa [31]2 years ago
3 0

Answer:

m_{Fe}=12.6gFe

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe

Best regards.

IceJOKER [234]2 years ago
3 0

Answer:

12.6g of Fe.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe2O3(s) —> 4Fe(s) + 3O2(g)

Next, we shall determine the mass of Fe2O3 that decomposed and the mass of Fe that is produced from the balanced equation.

This is illustrated below:

Mola mass of Fe2O3 = (56 x 2) + (16x3) = 160g/mol

Mass of Fe2O3 from the balanced equation = 2 x 160= 320g

Molar mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 4 x 56 = 224g

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Finall, we can obtain the mass of the Fe produced from the decomposition of 18g of Fe2O3 as follow:

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Therefore 18g of Fe2O3 will decompose to produce = (18x224)/320 = 12.6g of Fe.

Therefore, 12.6g of Fe is obtained from 18g of Fe2O3.

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5 0
3 years ago
A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample of the mixture is dissolved in water, and an ex- cess of NaOH is add
chubhunter [2.5K]

Answer:

The mass percent of aluminum sulfate in the sample is 16.18%.

Explanation:

Mass of the sample = 1.45 g

Al_2SO_3+6NaOH\rightarrow 2Al(OH)_3+3Na_2SO_4

Mass of the precipitate = 0.107 g

Moles of aluminum hydroxide = \frac{0.107 g}{78 g/mol}=0.001372 mol

According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .

Then 0.001372 moles of aluminum hydroxide will be obtained from:

\frac{1}{2}\times 0.001372 mol=0.000686 mol

Mass of 0.000686 moles of aluminum sulfate :

= 0.000686 mol × 342 g/mol = 0.2346 g

The mass percent of aluminum sulfate in the sample:

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5 0
3 years ago
What is the formula for tin(IV) sulfide?<br> A. Sn4S<br> B. SnS2<br> C. Sns<br> D. SnS4
defon

Answer:

SnS_{2}

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6 0
2 years ago
What is the H* concentration in a solution with a pH of 1.25?​
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Answer:

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Explanation:

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7 0
3 years ago
Determine the empirical formula of a compound containing 1.71 g of silicon and 8.63 g of chlorine.
Basile [38]

Answer:

The answer to your question is: SiCl₄

Explanation:

Data

amount of Si      1.71 g

amount of Cl     8.63 g

MW Si = 28 g

MW Cl = 35.5

Process (rule of three)

For Si                                                        For Cl

        28 g of Si ------------------ 1 mol                      35.5 g of Cl --------------- 1 mol

          1.71g of Si  ---------------   x                              8.63 g of Cl --------------  x

         x = 1.71 x 1 / 28 = 0.06 mol                          x = 8.63 x 1 / 35.5 = 0.24 mol

Now, divide both results by the lowest of them.

Si = 0.06 mol / 0.06 = 1 molecule of Si     Cl = 0.24 / 0.06 = 4 molecules of Cl

Finally

                     Si₁ Cl₄ or SiCl₄

8 0
3 years ago
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