Which of the following is not an example of temperature abuse ?

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

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

2 years ago

Based on the tiles shown, how many moles of oxygen gas (o2) are needed to produce 36.04 grams of water (h2o) when reacting with

madreJ [45]

We calculate for the number of moles of water given its mass by dividing the given mass by the molar mass.
n water = (36.04 g) / (18 g/mol)
n water = 2 mols
From the given balanced equation, every 6 moles of water produced will require 7 moles of oxygen.
n oxygen = (2 mols H2O) x (7 moles O2 / 6 moles H2O)
n oxygen = 2.33 mols O2

3 0

2 years ago

The vapor pressure of benzene, c6h6, is 100.0 torr at 26.1 °c. Assuming raoult's law is obeyed, how many moles of a nonvolatile

frutty [35]

According to Raoult's law:

$p_{solution} = X_{solvent}\times P_{solvent}$ -(1)

where $p_{solution}$ is observed vapor pressure of the solution, $X_{solvent}$ is mole fraction of solvent, and $P_{solvent}$ is vapor pressure of the pure solvent.

Vapor pressure of benzene = $100 torr at 26.1^{o} C$ (given)

Reduction in vapor pressure on addition of non-volatile solute = 10.0 % (given)

So, vapor pressure of the component in the solution = $100 - 10 = 90 torr$

Substituting the values in formula (1):

$90 = mole fraction of benzene \times 100$

$mole fraction of benzene = \frac{90}{100} = 0.9$

Mole of benzene = $Volume\times \frac{density}{Molar Mass}$

Mole of benzene = $100 cm^{3}\times \frac{0.8765 g/cm^{3}}{78 g/mol} = 1.124 mol$

Since, $mole fraction of benzene =\frac{mole of benzene }{mole of benzene +mole of non volatile solute }$

So, $0.9 =\frac{1.124 }{1.124+mole of non volatile solute }$

$\frac{1.124}{0.9} =1.124+mole of non volatile solute$

$mole of non volatile solute = 1.249 - 1.124 = 0.125$

Hence, moles of a nonvolatile solute to be added is $0.125$ .

3 0

2 years ago

A sample of a certain barium chloride hydrate, BaCl2.nH2O, has a mass of 7.62 g. When this sample is heated in a crucible, it is

Ratling [72]

Answer:

BaCl2.2H2O

Explanation:

Data obtained from the question include:

Mass of barium chloride hydrate (BaCl2.nH2O) = 7.62g

Mass of anhydrous barium chloride (BaCl2) = 6.48g

Next, we shall determine the mass of water in the barium chloride hydrate (BaCl2.nH2O). This can be obtained as follow:

Mass of water (H2O) = Mass of barium chloride hydrate (BaCl2.nH2O) – Mass of anhydrous barium chloride (BaCl2)

Mass of water (H2O) = 7.62 – 6.48

Mass of water (H2O) = 1.14g

Next, we shall write the balanced equation for the reaction. This is given below:

BaCl2.nH2O —> BaCl2 + nH2O

Next, we shall determine the masses of BaCl2.nH2O that decomposed and the mass H2O produced from the balanced equation.

Molar mass of BaCl2.nH2O = 137 + (35.5x2) + n[(2x1) + 16]

= 137 + 71 + n[2 + 16]

= (208 + 18n) g/mol

Mass of BaCl2.nH2O from the balanced equation = 1 x (208 + 18n) = (208 + 18n) g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = n x 18 = 18n g

Summary:

From the balanced equation above,

(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.

Finally, we shall determine the formula for the hydrated compound as follow:

From the balanced equation above,

(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.

But 7.62g of BaCl2.nH2O decomposed to produce 1.14g of H2O i.e

18n/(208 + 18n) = 1.14/7.62

Cross multiply

18n x 7.62 = 1.14(208 + 18n)

137.16n = 237.12 + 20.52n

Collect like terms

137.16n – 20.52n = 237.12

116.64n = 237.12

Divide both side by the coefficient of n i.e 116.64

n = 237.12/116.64

n = 2

Therefore the formula for the hydrate, BaCl2.nH2O is BaCl2.xH2O.

3 0

2 years ago