Question

1)к оксиду хрома (III) массой 19г добавлен алюминий . какая масса хрома образовалась, если массовая доля выхода его составила 90%?

Answer 1

Cr2O3+2Al=2Cr+Al2O3

 

n(Cr2O3)=19/152=0,125 моль

m(Cr)=0,25*0,9*52=11.7г

Answer 2

Calculating the Mass of Chromium (III) Oxide Formed

To calculate the mass of chromium formed when aluminum is added to chromium (III) oxide, we can use the concept of stoichiometry and the given information about the mass of chromium (III) oxide and the mass fraction of its yield.

Given: - Mass of chromium (III) oxide = 19g - Mass fraction of the yield = 90%

We can start by determining the molar mass of chromium (III) oxide and the stoichiometry of the reaction between aluminum and chromium (III) oxide.

The molar mass of chromium (III) oxide (Cr2O3) can be calculated using the individual molar masses of chromium and oxygen: - Molar mass of chromium (Cr) = 51.996 g/mol - Molar mass of oxygen (O) = 15.999 g/mol

The molar mass of chromium (III) oxide (Cr2O3) is: Molar mass of Cr2O3 = (2 * molar mass of Cr) + (3 * molar mass of O)

Now, let's calculate the molar mass of chromium (III) oxide.

Molar mass of Cr2O3 = (2 * 51.996 g/mol) + (3 * 15.999 g/mol)

Molar mass of Cr2O3 = 151.996 g/mol

Next, we need to determine the stoichiometry of the reaction between aluminum and chromium (III) oxide to find the amount of chromium formed.

The balanced chemical equation for the reaction is: 2Al + Cr2O3 → 2Al2O3 + 2Cr

From the balanced chemical equation, we can see that 1 mole of chromium (III) oxide (Cr2O3) produces 2 moles of chromium (Cr).

Now, we can calculate the amount of chromium formed using the given mass of chromium (III) oxide and the mass fraction of its yield.

The mass of chromium formed can be calculated using the following steps: 1. Calculate the number of moles of chromium (III) oxide. 2. Determine the theoretical yield of chromium. 3. Calculate the actual yield of chromium. 4. Determine the mass of chromium formed.

Let's proceed with the calculations based on the provided information.

Calculations

1. Number of moles of chromium (III) oxide: - Number of moles = Mass / Molar mass - Number of moles = 19g / 151.996 g/mol

2. Theoretical yield of chromium: - Theoretical yield of chromium = Number of moles of Cr2O3 * 2 moles of Cr / 1 mole of Cr2O3

3. Actual yield of chromium: - Actual yield of chromium = Theoretical yield * Mass fraction of the yield

4. Mass of chromium formed: - Mass of chromium formed = Actual yield of chromium * Molar mass of Cr

Let's calculate the mass of chromium formed using the above steps.

Results

1. Number of moles of chromium (III) oxide: - Number of moles = 19g / 151.996 g/mol - Number of moles ≈ 0.125 moles

2. Theoretical yield of chromium: - Theoretical yield of chromium = 0.125 moles * 2 moles / 1 mole - Theoretical yield of chromium ≈ 0.25 moles

3. Actual yield of chromium: - Actual yield of chromium = 0.25 moles * 90% - Actual yield of chromium ≈ 0.225 moles

4. Mass of chromium formed: - Mass of chromium formed = 0.225 moles * 51.996 g/mol - Mass of chromium formed ≈ 11.699 g

Conclusion

The mass of chromium formed, when aluminum is added to chromium (III) oxide and the mass fraction of the yield is 90%, is approximately 11.699 grams.