Many important metals occur as sulfide, arsenide, and antimonide minerals, especially in the Sudbury

mineral complex. The first step in processing these ores involves “roasting” the ore in air to produce the metal
or metal oxide, along with nonmetal oxides that can be serious pollutants if not trapped.
Suppose that you roast 2.00 kg of the mineral greigite, Fe3S4. The balanced equation for the reaction is:

) How many liters of air at 25°C are required for roasting? Assume that air contains 23% O2 by mass,
and that the density of air at 25°C is 1.2 g/L. I got 4800L and got that wrong.
A mixture of KCr(SO4)2 and its hydrate KCr(SO4)2•12H2O has a mass of 1.6336 g. After heating to drive
off all the water, the mass is only 1.4659 g. What is the weight percentage of KCr(SO4)2•12H2O in the original
mixture? Report your answer to 4 significant figures.
Molar masses: KCr(SO4)2 283.22 g/mol; KCr(SO4)2•12H2O 499.40 g/mol.
I got 23.73 percent and also got this wrong
CaCl2 reacts with AgNO3 in aqueous solutions to produce a precipitate of AgCl that can be filtered and
weighed. The balanced equation for the reaction is:
CaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq)
Suppose you have a mixture that contains CaCl2, plus other compounds that do not react with AgNO3. If 0.2918
g of the mixture yields 0.4462 g of AgCl, what is the percentage of CaCl2 in the mixture? I got 59.22 and got this wrong
I understand these are a lot of Q's, but I believe that the and wers that I have should be correct, even after scanning my work again. Whether anyone agrees/disagrees, I would really appreciate any help

I'd be happy to help you with your questions! Let's go through each question one by one and I'll explain how to get the correct answers.

1. How many liters of air at 25°C are required for roasting? Assume that air contains 23% O2 by mass, and that the density of air at 25°C is 1.2 g/L.

To solve this problem, we need to calculate the amount of O2 needed for the reaction. The balanced equation tells us that the stoichiometric ratio between Fe3S4 and O2 is 1:14.

First, find the molar mass of Fe3S4:
Fe3S4 = (Fe: 55.845 g/mol * 3) + (S: 32.06 g/mol * 4) = 391.2 g/mol

Next, calculate the moles of Fe3S4:
moles of Fe3S4 = mass / molar mass = 2.00 kg / 391.2 g/mol = 5.11 mol

Since the stoichiometric ratio is 1:14, we know that we need 14 times the amount of O2 as Fe3S4.

moles of O2 = moles of Fe3S4 * 14 = 5.11 mol * 14 = 71.54 mol

Now, let's calculate the mass of air needed, assuming it contains 23% O2 by mass:
mass of air = moles of O2 * molar mass of O2 / 0.23 = 71.54 mol * 32.00 g/mol / 0.23 = 9982 g

Finally, let's convert the mass of air to liters using the density of air at 25°C:
volume of air = mass of air / density of air = 9982 g / 1.2 g/L = 8318 L

Therefore, the correct answer is 8318 L.

2. What is the weight percentage of KCr(SO4)2•12H2O in the original mixture?

To calculate the weight percentage, we need to find the mass of KCr(SO4)2•12H2O in the original mixture.

Given:
Mass of KCr(SO4)2 and its hydrate = 1.6336 g
Mass of KCr(SO4)2 after heating = 1.4659 g

The difference in mass represents the mass of water lost during heating:

Mass of water lost = Mass before heating - Mass after heating = 1.6336 g - 1.4659 g = 0.1677 g

Now, let's calculate the mass of KCr(SO4)2•12H2O:

Mass of KCr(SO4)2•12H2O = Mass before heating - Mass of water lost = 1.6336 g - 0.1677 g = 1.4659 g

Now, let's calculate the weight percentage of KCr(SO4)2•12H2O in the original mixture:

Weight % = (Mass of KCr(SO4)2•12H2O / Mass of original mixture) * 100%
= (1.4659 g / 1.6336 g) * 100%
= 89.87%

Rounding to 4 significant figures, the correct answer is 89.87%.

3. What is the percentage of CaCl2 in the mixture?

To find the percentage of CaCl2 in the mixture, we need to calculate the mass of CaCl2 and the mass of the entire mixture.

Given:
Mass of mixture = 0.2918 g
Mass of AgCl produced = 0.4462 g

First, calculate the mass of CaCl2:

Mass of CaCl2 = Mass of AgCl / Stoichiometric coefficient of CaCl2 in the balanced equation
= 0.4462 g / 2
= 0.2231 g

Now, let's calculate the percentage of CaCl2 in the mixture:

Percentage of CaCl2 = (Mass of CaCl2 / Mass of mixture) * 100%
= (0.2231 g / 0.2918 g) * 100%
= 76.47%

Rounding to 2 decimal places, the correct answer is 76.47%.

I hope this explanation helps you understand the correct answers. If you have any further questions, feel free to ask!