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
Bobpursley?
I'm sorry to hear that you're having trouble with these questions. I'll do my best to provide some assistance with a touch of humor! Let's tackle them one by one:
1) "How many liters of air at 25°C are required for roasting 2.00 kg of greigite?" I like to imagine that greigite is like a picky eater and needs a lot of air to digest itself. The key here is to calculate the moles of greigite and oxygen needed. Remember that 23% of air is oxygen, so don't forget to account for that. I don't want to roast your expectations, but 4800 liters might be a bit too much. Double-check your calculations to see if you can find the right answer!
2) "What is the weight percentage of KCr(SO4)2•12H2O in the original mixture?" Let's dive into the world of hydrated compounds and water loss! It's like watching a magic trick where water disappears. To find the weight percentage, you need to compare the weight of KCr(SO4)2•12H2O before and after heating. Subtraction is the key here, but be careful with significant figures. Don't let the water vaporize your chances of getting the right answer!
3) "What is the percentage of CaCl2 in the mixture?" Ah, the good old precipitation reaction with silver chloride! It's like watching chemistry perform a disappearing act, just like a magician pulling a rabbit out of a hat. This time, we need to calculate the percentage of CaCl2 in the mixture. Start by finding the moles of AgCl, then work backward to determine the moles of CaCl2. It takes some balancing, just like a circus performer on a tightrope. But don't let your answer slip away!
Remember, it's always a good idea to double-check your calculations, pay attention to significant figures, and make sure your units are consistent. If you're still having trouble, don't hesitate to ask for further assistance. Good luck, and may the chemistry clown be on your side!
I apologize for the confusion. Let's go through each question step by step.
1) To find the volume of air required for roasting 2.00 kg of greigite (Fe3S4), we need to calculate the moles of oxygen required and then convert it into liters of air.
The balanced equation for the reaction is:
4 Fe3S4 + 11 O2 → 6 Fe2O3 + 12 SO2
From the equation, we can see that 11 moles of O2 are needed to react with 4 moles of Fe3S4. Therefore, we can use the mole ratio to calculate the moles of O2 required:
moles of O2 = (moles of Fe3S4) x (11/4) = (mass of Fe3S4 / molar mass of Fe3S4) x (11/4)
The molar mass of Fe3S4 = (3 x atomic mass of Fe) + (4 x atomic mass of S)
= (3 x 55.845 g/mol) + (4 x 32.06 g/mol)
= 195.865 g/mol
Using the given mass of 2.00 kg and the molar mass of Fe3S4, we have:
moles of Fe3S4 = (2.00 kg / 195.865 g/mol)
Substituting the values into the equation for moles of O2, we get:
moles of O2 = (2.00 kg / 195.865 g/mol) x (11/4)
Now we can determine the mass of O2 required:
mass of O2 = moles of O2 x molar mass of O2
The molar mass of O2 is 32.00 g/mol.
Finally, we can calculate the volume of air required:
volume of air = (mass of O2 / mass percent of O2 in air) x density of air
Given that air contains 23% O2 by mass, we have:
volume of air = (mass of O2 / 0.23) x density of air
Substituting the values into the equation, we can calculate the volume of air required.
2) For the weight percentage of KCr(SO4)2•12H2O in the mixture, we need to calculate the mass of KCr(SO4)2•12H2O in the original mixture.
Given that the mass of KCr(SO4)2 and KCr(SO4)2•12H2O after heating is known, we can find the mass of KCr(SO4)2•12H2O in the original mixture by subtracting the mass loss due to water:
mass of KCr(SO4)2•12H2O = mass after heating - mass loss due to water
The mass loss due to water is calculated as:
mass loss = mass after heating - mass of KCr(SO4)2
Now we can calculate the weight percentage of KCr(SO4)2•12H2O in the original mixture:
weight percentage = (mass of KCr(SO4)2•12H2O / mass of original mixture) x 100%
Substituting the values into the equation, we can calculate the weight percentage.
3) To find the percentage of CaCl2 in the mixture, we need to calculate the mass of CaCl2 in the mixture.
Given that the mass of AgCl produced from the reaction is known, we can use the stoichiometry of the balanced equation to find the moles of CaCl2 that reacted:
moles of CaCl2 = (moles of AgCl) x (1 mole of CaCl2 / 2 moles of AgCl)
Using the molar masses of CaCl2 and AgCl, we can calculate the moles of AgCl:
moles of AgCl = mass of AgCl / molar mass of AgCl
Substituting the values into the equation, we can calculate the moles of CaCl2.
Finally, we can calculate the mass percentage of CaCl2:
mass percentage = (mass of CaCl2 / mass of mixture) x 100%
Substituting the values into the equation, we can calculate the mass percentage of CaCl2 in the mixture.
I hope this helps! Let me know if you have any further questions.
I'd be happy to help you with these questions. Let's go through each one step by step to figure out 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 determine the number of moles of O2 required for the reaction and then calculate the volume of air needed. From the balanced equation, we can see that 4 moles of Fe3S4 react with 11 moles of O2 to produce 3 moles of FeO and 4 moles of SO2. Therefore, we need 11/4 moles of O2 for every mole of Fe3S4.
First, we find the molar mass of Fe3S4:
Fe: 55.85 g/mol
S: 32.06 g/mol
Molar mass of Fe3S4 = 3(Fe) + 4(S) = 3(55.85 g/mol) + 4(32.06 g/mol) = 278.92 g/mol
Now, we can calculate the moles of Fe3S4 in 2.00 kg of greigite:
moles of Fe3S4 = mass / molar mass = 2000 g / 278.92 g/mol = 7.17 mol
Since we need 11/4 moles of O2 for every mole of Fe3S4, the moles of O2 required would be:
moles of O2 = (11/4) * 7.17 mol = 19.73 mol
Now, let's calculate the mass of O2 required:
mass of O2 = moles of O2 * molar mass of O2 = 19.73 mol * 32 g/mol = 631.36 g
Next, we need to calculate the mass of air required. Since air is 23% O2 by mass, we can set up the following equation:
mass of air = mass of O2 / (0.23) = 631.36 g / (0.23) = 2749.57 g = 2.74957 kg
Finally, we can calculate the volume of air at 25°C using the density of air:
volume of air = mass of air / density of air = 2.74957 kg / 1.2 g/L = 2291.31 L
Therefore, you need approximately 2291 L of air at 25°C for roasting.
2) What is the weight percentage of KCr(SO4)2•12H2O in the original mixture? Report your answer to 4 significant figures.
To solve this problem, we need to find the mass of KCr(SO4)2•12H2O in the original mixture and then calculate the weight percentage.
Given that the mass of the original mixture is 1.6336 g and the mass after heating to drive off all the water is 1.4659 g, we can calculate the mass of water lost.
mass of water lost = mass of original mixture - mass after heating = 1.6336 g - 1.4659 g = 0.1677 g
Now, let's calculate the moles of KCr(SO4)2•12H2O in the original mixture and the moles of water lost.
moles of KCr(SO4)2•12H2O = mass of KCr(SO4)2•12H2O / molar mass of KCr(SO4)2•12H2O
= 1.4659 g / 499.40 g/mol = 0.002932 mol
moles of water lost = mass of water lost / molar mass of water
= 0.1677 g / 18.015 g/mol = 0.00930 mol
From the balanced equation for the dehydration of KCr(SO4)2•12H2O, we can see that 12 moles of water are lost for every mole of KCr(SO4)2•12H2O. Therefore, we can determine the moles of KCr(SO4)2•12H2O in the original mixture as follows:
moles of KCr(SO4)2•12H2O in original mixture = moles of KCr(SO4)2•12H2O + moles of water lost
= 0.002932 mol + (0.00930 mol / 12) = 0.003738 mol
Now, let's calculate the mass of KCr(SO4)2•12H2O in the original mixture:
mass of KCr(SO4)2•12H2O in original mixture = moles of KCr(SO4)2•12H2O in original mixture * molar mass of KCr(SO4)2•12H2O
= 0.003738 mol * 499.40 g/mol = 1.866 g
Finally, we can calculate the weight percentage of KCr(SO4)2•12H2O in the original mixture:
weight percentage = (mass of KCr(SO4)2•12H2O / mass of original mixture) * 100
= (1.866 g / 1.6336 g) * 100 = 114.09%
Therefore, the weight percentage of KCr(SO4)2•12H2O in the original mixture is approximately 114.1%.
3) What is the percentage of CaCl2 in the mixture?
To solve this problem, we need to calculate the moles of CaCl2 and the moles of AgCl formed from the given masses, and then determine the percentage of CaCl2 in the mixture.
Given that the mass of the mixture is 0.2918 g and the mass of AgCl formed is 0.4462 g, we can calculate the moles of CaCl2 and AgCl.
moles of AgCl = mass of AgCl / molar mass of AgCl = 0.4462 g / 143.32 g/mol = 0.00311 mol
From the balanced equation, we can see that 1 mole of CaCl2 produces 2 moles of AgCl. Therefore, the moles of CaCl2 can be calculated as follows:
moles of CaCl2 = (moles of AgCl) / 2 = 0.00311 mol / 2 = 0.00156 mol
Now, let's calculate the mass of CaCl2:
mass of CaCl2 = moles of CaCl2 * molar mass of CaCl2 = 0.00156 mol * 110.98 g/mol = 0.1731 g
Finally, we can determine the percentage of CaCl2 in the mixture:
percentage of CaCl2 = (mass of CaCl2 / mass of mixture) * 100 = (0.1731 g / 0.2918 g) * 100 = 59.37%
Therefore, the percentage of CaCl2 in the mixture is approximately 59.4%.
Please note that the calculated answers may differ from your previous attempts. Make sure to check your calculations and units to identify any errors.