I have multiple...

1) If a metal ore contains 23.1% pure metal, how
many pounds of pure metal can be obtained
from 38 pounds of metal ore?
Answer in units of pounds.

2) For the reaction
3 F2 + Cl2 → 2 ClF3 ,
what is the maximum amount of ClF3
(92.4482 g/mol) which could be formed from
15.83 g of F2 (37.9968 g/mol) and 0.264 mol
of Cl2 (70.906 g/mol)?
Answer in units of mol.

- I already balanced the equation. My previous incorrect answers were 0.27, 0.528, and 0.53.

3) An ore contains Fe3O4 and no other iron. The
iron in a 56-gram sample of the ore is all
converted by a series of chemical reactions to
Fe2O3. The mass of Fe2O3 is measured to be
21.2 g. What was the mass of Fe3O4 in the
sample of ore?
Answer in units of g.

- Previous incorrect answers: 30.74, 40.52, 15.34, and 14.83.

4) The reaction of 9.9 grams of fluorine with
excess chlorine produced 3.3 grams of ClF3.
What percent yield of ClF3 was obtained?
Answer in units of %.

- Previous incorrect answers: 35.55 and 10.28

5) A 26.4501 g sample of impure magnesium carbonate
was heated to complete decomposition
according to the equation
MgCO3(s) → MgO(s) + CO2(g).
After the reaction was complete, the solid
residue (consisting of MgO and the original
impurities) had a mass of 17.435 g. Assuming
that only the magnesium carbonate had decomposed,
how much magnesium carbonate
was present in the original sample?
Answer in units of g.

6) An ore contains Fe3O4 and no other iron. The
iron in a 40.93 gram sample of the ore is all
converted by a series of chemical reactions to
Fe2O3. The mass of Fe2O3 is measured to be
30.7 grams. What was the percent Fe3O4 in
the sample of ore?
Answer in units of %.

- Previous incorrect answer: 52.46

Let's go step-by-step to solve each of these questions:

1) To find the amount of pure metal obtained from a metal ore, you need to multiply the weight of the metal ore by the percentage of pure metal it contains. In this case, the percentage is given as 23.1%:

Pure metal = Weight of metal ore * Percentage of pure metal
Pure metal = 38 pounds * 23.1% = 8.778 pounds

Therefore, you can obtain 8.778 pounds of pure metal from 38 pounds of metal ore.

2) To find the maximum amount of a product formed in a chemical reaction, you need to use the concept of limiting reactant. First, convert the given masses of F2 and Cl2 to moles using their molar masses:

Moles of F2 = Mass of F2 / Molar mass of F2
Moles of F2 = 15.83 g / 37.9968 g/mol = 0.416 mol

Moles of Cl2 = 0.264 mol (given)

Next, compare the moles of F2 and Cl2 using the balanced equation coefficients. The balanced equation shows that the ratio of F2 to Cl2 is 3:1, which means 3 moles of F2 react with 1 mole of Cl2 to produce 2 moles of ClF3.

Since the ratio of F2 to Cl2 is 3:1 and there is excess Cl2, F2 is the limiting reactant. Therefore, we can use the moles of F2 to calculate the maximum amount of ClF3 formed:

Moles of ClF3 = (Moles of F2 * 2) = (0.416 mol * 2) = 0.832 mol

The maximum amount of ClF3 that can be formed is 0.832 mol.

3) To find the mass of Fe3O4 in the sample of ore, you need to determine the mass of Fe3O4 that was converted to Fe2O3.

First, calculate the molar mass of Fe3O4:
Molar mass of Fe3O4 = (3 * Molar mass of Fe) + (4 * Molar mass of O)
Molar mass of Fe3O4 = (3 * 55.845 g/mol) + (4 * 16.00 g/mol)
Molar mass of Fe3O4 = 159.69 g/mol

Next, calculate the moles of Fe2O3 formed:
Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
Moles of Fe2O3 = 21.2 g / 159.69 g/mol
Moles of Fe2O3 = 0.1329 mol

Since the reaction is balanced such that 1 mole of Fe3O4 produces 1 mole of Fe2O3, the moles of Fe3O4 in the sample is also 0.1329 mol.

To find the mass of Fe3O4, use its molar mass:
Mass of Fe3O4 = Moles of Fe3O4 * Molar mass of Fe3O4
Mass of Fe3O4 = 0.1329 mol * 159.69 g/mol
Mass of Fe3O4 = 21.21 g

Therefore, the mass of Fe3O4 in the sample of ore is 21.21 g.

4) To calculate the percent yield of a reaction, divide the actual yield by the theoretical yield and multiply by 100%. The actual yield is given as 3.3 grams and the theoretical yield is what could be obtained based on the balanced equation and stoichiometry.

First, calculate the theoretical yield of ClF3:

Moles of ClF3 = Mass of ClF3 / Molar mass of ClF3
Moles of ClF3 = 3.3 g / 92.4482 g/mol
Moles of ClF3 = 0.0357 mol (approx.)

The stoichiometry of the reaction shows that 3 moles of F2 produce 2 moles of ClF3. Since F2 is the limiting reactant, we compare the moles of ClF3 to the moles of F2 (0.416 mol):

Ratio of ClF3 to F2 = (0.0357 mol / 0.416 mol) = 0.086

This means that in theory, 0.086 moles of ClF3 could be formed from the given reactants.

Now we can calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) * 100%
Percent yield = (3.3 g / 0.086 mol) * 100%
Percent yield = 3837% (approx.)

Therefore, the percent yield of ClF3 is approximately 3837%.

5) To find the amount of magnesium carbonate present in the original sample, you need to find the mass of the magnesium carbonate that decomposed. Since the reaction is balanced such that 1 mole of MgCO3 produces 1 mole of MgO, we can calculate the moles of MgO formed:

Moles of MgO = Mass of MgO / Molar mass of MgO
Moles of MgO = 17.435 g / 40.30 g/mol (molar mass of MgO)
Moles of MgO = 0.432 mol (approx.)

Since 1 mole of MgCO3 produces 1 mole of MgO, the moles of MgCO3 present originally would be the same, so:

Mass of MgCO3 = Moles of MgCO3 * Molar mass of MgCO3
Mass of MgCO3 = 0.432 mol * 84.31 g/mol (molar mass of MgCO3)
Mass of MgCO3 = 36.43 g (approx.)

Therefore, the mass of magnesium carbonate present in the original sample is approximately 36.43 g.

6) To find the percent Fe3O4 in the sample of ore, we need to compare the mass of Fe3O4 to the total mass of the sample.

First, calculate the mass of Fe3O4 using stoichiometry:

Moles of Fe3O4 = Mass of Fe3O4 / Molar mass of Fe3O4
Moles of Fe3O4 = 30.7 g / 159.69 g/mol
Moles of Fe3O4 = 0.1923 mol

Since the balanced equation shows that 1 mole of Fe3O4 produces 1 mole of Fe2O3, the moles of Fe3O4 in the sample is 0.1923 mol.

Next, calculate the total mass of the sample:

Total mass of the sample = Mass of Fe3O4 + Mass of Fe2O3
Total mass of the sample = 30.7 g + 40.93 g = 71.63 g

Now we can calculate the percent Fe3O4 in the sample:

Percent Fe3O4 = (Mass of Fe3O4 / Total mass of the sample) * 100%
Percent Fe3O4 = (30.7 g / 71.63 g) * 100%
Percent Fe3O4 = 42.85%

Therefore, the percent Fe3O4 in the sample of ore is approximately 42.85%.