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Olegator [25]
3 years ago
15

How many grams of CO are needed to react with an excess of Fe2O3 to produce (210.3 g Fe) 209.7 g Fe? Show your work.

Chemistry
2 answers:
Flura [38]3 years ago
6 0
The balanced chemical reaction is given as:

<span>3CO (g) + Fe2O3 (s) --> 2Fe (s) + 3CO2 (g)

We are given the amount of the product to be produced from the reaction. We use this as the starting value for the calculations. We do as follows:

209.7 g Fe ( 1 mol / 210.3 g Fe ) ( 3 mol CO / 2 mol Fe ) (28.01 g / 1 mol) = 41.90 g CO needed</span>
Fed [463]3 years ago
4 0

Explanation:

3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of Fe in 210.3 g=

\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}

=\frac{210.3}{55.84 g/mol}=3.76 mol

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : \frac{3}{2}\times 3.76 moles of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of Fe in 209.7 g=

\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}

=\frac{209.7}{55.84 g/mol}=3.75 mol

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : \frac{3}{2}\times 3.75 moles of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g

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5 0
3 years ago
Read 2 more answers
A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

Next, the total moles:

n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0
3 years ago
Develop a demonstration to show how weight is not the same thing as mass
maksim [4K]

Answer with Explanation:

"Mass" and "weight" should never be used interchangeably with each other. Mass refers to the <u>total amount of matter</u><u> that can be measured in an object, </u>while weight refers to the<u> measure of the</u><u> force of gravity</u><u> that is acting on the object's mass.</u>

The mass of an object is<u> constant</u> (meaning, it doesn't change even if the object will be placed on another location) while the weight of an object relies on the <em>force of gravity.</em> So, this means that your mass on Earth and on the moon are identical, however, your weight on Earth and on the Moon are different. You will weigh lesser on the Moon because it has a lesser surface gravity than that of Earth.

So, this explains the answer.

4 0
3 years ago
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