All categories

2 years ago

What mass of oxygen reacts when 84.9 g of iron is consumed in the following reaction: Fe+O2= Fe2O3

2 answers:

6 0

First we will calculate the number of moles of Iron:
$n = \frac{m}{M}
$ , where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
$n= \frac{84.9}{55.845} = 1,52$ moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So $1,52= \frac{mOxygen}{32}$ where 32 is the Atomical Weight of Oxygen (16 x 2).
=> $mOxygen=32*1,52=48,64$ g

saul85 [17]2 years ago

3 0

Answer:

36.385 grams of oxygen reacts when 84.9 grams of iron.

Explanation:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Moles of iron = $\frac{84.9 g}{56 g/mol}=1.5160 mol$

According to reaction, 4 moles of iron reacts with 3 moles of oxygen gas.

Then 1.5160 moles of iron will react with:

$\frac{3}{4}\times 1.5160 mol=1.1370 mol$ of oxygen gas

Mass of 1.1370 moles of oxygen gas:

$1.1370 mol\times 32 g/mol=36.385 g$

36.385 grams of oxygen reacts when 84.9 grams of iron.

You might be interested in

Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod

son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0

2 years ago

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HC

SVETLANKA909090 [29]

Answer:

The correct answer is is option B

b. 93.3 g

Explanation:

SEE COMPLETE QUESTION BELOW

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)

How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?

a. 7.30 g

b. 93.3 g

c. 146 g

d. 150 g

e. 196 g

CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION

7 0

2 years ago

A student dissolved a sample in hexane, spotted it on to a TLC plate and eluted using ethyl acetate. After visualizing the TLC p

Zepler [3.9K]

1) Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:

Determine the Rf values of compounds A, B, and C run on a silica gel TLC plate using hexanes as the solvent.Which compound, A, B, or C, is the most polar?What would you expect to happen to the Rf values if you used acetone instead of hexanes as the eluting solvent?How would the Rf values change if eluted with hexanes using an alumina TLC plate?

Show answer

2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?

Show answer

3) After a rather lengthy organic chemistry synthesis procedure, a student ran the product of the reaction on a TLC plate and obtained the result below. What might he/she have done wrong, if anything?

Show answer

4) A student spots an unknown sample on a TLC plate. After developing in hexanes/ethyl acetate 50:50, he/she saw a single spot with an Rf of 0.55. Does this indicate that the unknown material is a pure compound? What can be done to verify the purity of the sample?

Show answer

5) Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC plate and developed in a hexanes/ethyl acetate solvent mixture. Predict the relative Rf values for the three compounds in the sample.

Show answer

6) Plate A, below, represents the TLC chromatogram of a compound run in hexanes. The same compound was then spotted on a large TLC plate and again run in hexanes. Which TLC plate, B, C, or D, correctly represents how far the compound would run on the longer plate?

Show answer

Back to TLC

Original content © University of Colorado at Boulder, Department of Chemistry and Biochemistry.
The information on these pages is available for academic use without restriction.

3 0

2 years ago

A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

ioda

In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = $\frac{PV}{nT}$ = constant
$\frac{P1 V1}{n T}$ = $\frac{P2 V2}{n T}$
==> P1 * V1 = P2 * V2
V2 = $\frac{P1 V1}{P2}$ = $\frac{445 0.016}{1225}$
V2 = 0.00581 m3 = 5.81 L

7 0

2 years ago

How many grams of Cr can be produced by the reaction of 44.1 g of Cr2O3 with 35.0 g of Al according to the following chemical eq

allsm [11]

Answer:

30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.

19.33 grams of the excess react will remain once the reaction goes to completion.

Explanation:

You know:

2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of the compounds participating in the reaction react and are produced:

Al: 2 moles
Cr₂O₃: 1 mole
Al₂O₃: 1 mole
Cr: 2 moles

Being:

Al: 27 g/mole
Cr: 52 g/mole
O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

Al: 27 g/mole
Cr₂O₃: 2*52 g/mole + 3 *16 g/mole= 152 g/mole
Al₂O₃: 2*27 g/mole + 3 *16 g/mole= 102 g/mole
Cr: 52 g/mole

Then, by stoichiometry of the reaction, the amounts of reagents and products that participate in the reaction are:

Al: 2 moles* 27 g/mole= 54 g
Cr₂O₃: 1 mole* 152 g/mole= 152 g
Al₂O₃: 1 mole* 102 g/mole= 102 g
Cr: 2moles*52 g/mole= 104 g

First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

Then, for this you must apply the following rule of three: if 54 grams of Al react with 152 grams of Cr₂O₃ by stoichiometry of the reaction, 35 grams of Al with how much mass of Cr₂O₃ will react?

$mass of Cr_{2} O_{3} =\frac{35 grams of Al*152 gramsof Cr_{2} O_{3}}{54 grams of Al}$

mass of Cr₂O₃= 98.52 grams

But 98.52 grams of Cr₂O₃ are not available, 44.1 grams are available. Since you have less moles than you need to react with 35 grams of Al, the compound Cr₂O₃ will be the limiting reagent.

To determine the amount of Cr that can be produced, you use the amount of limiting reagent available and apply the following rule of three: if by stoichiometry 152 grams of Cr₂O₃ produce 104 grams of Cr, 44.1 grams of Cr₂O₃ how much mass of Cr does it produce?

$mass of Cr =\frac{104 grams of Cr*44.1 grams of Cr_{2} O_{3}}{152 grams of Cr_{2} O_{3}}$

mass of Cr= 30.17 grams

30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.

As Al is the excess reagent, you must first calculate the amount of mass that reacts with 44.1 grams of Cr₂O₃ using the following rule of three: if 152 grams of Cr₂O₃ react with 54 grams of Al, 44.1 grams of Cr₂O₃ with how much mass of Al reaction to?

$mass of Al =\frac{54 grams of Al*44.1 gramsnof Cr_{2} O_{3}}{152 gramsnof Cr_{2} O_{3}l}$

mass of Al=15.67 grams

With 35 grams being the amount of Al available, the amount of Al that will remain in the reaction after all the limiting reagent reacts and the reaction is complete is calculated by:

mass of excess= 35 grams - 15.67 grams= 19.33 grams

19.33 grams of the excess react will remain once the reaction goes to completion.

6 0

2 years ago