A certain mine produces 2.5 108 kg of copper from chalcopyrite (CuFeS2) each year. The ore contains only 0.80% Cu by mass.
(a) If the density of the ore is 2.0 g/cm3, calculate the volume of ore removed each year.
(b) Calculate the mass of SO2 produced by roasting (assume chalcopyrite to be the only source of sulfur).
I don't even know where to begin.
For a, 0.8% of some mass (say x) = 2.5108 kg. Calculate x = mass ore that must be mined to obtain 2.5 kg Cu.
Change that to grams, then mass = volume x density. You know mass and density; calculatr volume. Part b is a stoichiometry problem. Write the equation, balance it, convert g Cu to mols, convert mols Cu to mols SO2, convert mols SO2 to grams. Post any work if you need futher assistance.
Oh boy, here we go! Let's see if I can help you figure this out while keeping things light and fun.
(a) So, we need to calculate the volume of ore removed each year. Let's break it down step by step.
First, we know that the ore contains 0.80% Cu by mass. That means for every 100 grams of ore, 0.80 grams will be copper.
Now, we're given that the mine produces 2.5 x 10^8 kg of copper each year from the chalcopyrite. So, let's convert that to grams: 2.5 x 10^8 kg = 2.5 x 10^11 g.
Since we know that every 100 grams of ore contains 0.80 grams of copper, we can set up a proportion to calculate the mass of ore needed:
(0.80 g Cu / 100 g ore) = (2.5 x 10^11 g Cu / x g ore)
Now we can solve for x, which represents the mass of ore needed.
But wait, we're not done yet! We still need to calculate the volume of the ore. We're given that the density of the ore is 2.0 g/cm^3. So,
mass = volume x density
Plug in the values you found for mass and density, and you'll be able to calculate the volume of ore removed each year.
(b) Alright, let's move on to part b! We need to calculate the mass of SO2 produced by roasting, assuming that chalcopyrite is the only source of sulfur.
To do this, we'll use some good old stoichiometry.
First, let's write the balanced equation for the reaction:
CuFeS2 + O2 -> Cu + FeO + SO2
From the equation, we can see that for every 1 mol of Cu, we get 1 mol of SO2.
We know that the mass of Cu produced is 2.5 x 10^11 g (from part a). So, let's convert that to moles of Cu by dividing by the molar mass of Cu.
Then, using the mole ratio from the balanced equation, we can convert from moles of Cu to moles of SO2.
Finally, convert moles of SO2 to grams by multiplying by the molar mass of SO2.
And there you have it! You'll be able to find the mass of SO2 produced by roasting.
I hope that helps you get started on solving these problems. Remember, if you need any further assistance, just let me know!
(a) To find the volume of ore removed each year, we need to first calculate the mass of the ore mined to obtain 2.5 x 10^8 kg of copper (Cu).
Given that the ore contains only 0.80% Cu by mass, we can set up the following equation:
0.80% of mass = 2.5 x 10^8 kg
Let's solve for the mass:
0.80/100 * mass = 2.5 x 10^8 kg
mass = (2.5 x 10^8 kg) / (0.80/100)
mass = 3.125 x 10^10 kg
Now, we can calculate the volume using the mass and density:
density = mass / volume
Rearranging the equation, we have:
volume = mass / density
volume = (3.125 x 10^10 kg) / (2.0 g/cm^3)
Note: We need to convert the density to grams per cubic centimeter (g/cm^3) since the mass is in kilograms (kg).
volume = (3.125 x 10^10 kg) / (2.0 x 10^3 g/cm^3)
Now, we can calculate the volume:
volume ≈ 1.5625 x 10^7 cm^3
Therefore, the volume of ore removed each year is approximately 1.5625 x 10^7 cubic centimeters (cm^3).
(b) To calculate the mass of sulfur dioxide (SO2) produced by roasting, we first need to determine the moles of sulfur (S) present in the chalcopyrite (CuFeS2).
The balanced equation for the roasting process is as follows:
2CuFeS2(s) + 3O2(g) → 2FeO(s) + 2CuS(l) + 2SO2(g)
From the balanced equation, we can see that for every 2 moles of CuFeS2, we get 2 moles of CuS and 2 moles of SO2.
To find the moles of sulfur in CuFeS2, we can use the molar mass of sulfur (32.07 g/mol) and the mass of CuFeS2:
moles of S = mass of CuFeS2 / molar mass of S
moles of S = (3.125 x 10^10 kg) * (0.008) / 32.07 g/mol
Now, we can calculate the moles of SO2 produced:
moles of SO2 = moles of S
Finally, to find the mass of SO2 produced, we can use the molar mass of SO2 (64.07 g/mol):
mass of SO2 = moles of SO2 * molar mass of SO2
mass of SO2 = (3.125 x 10^10 kg) * (0.008) / 32.07 g/mol * 64.07 g/mol
Simplifying the equation, we have:
mass of SO2 ≈ 1.5625 x 10^7 g
Therefore, the mass of SO2 produced by roasting is approximately 1.5625 x 10^7 grams (g).
To solve part (a), we need to find the volume of the ore removed each year. We are given that the ore contains 0.80% Cu by mass and that the mine produces 2.5 * 10^8 kg of copper each year.
Let's start by calculating the mass of the ore that must be mined to obtain 2.5 * 10^8 kg of copper.
Step 1: Set up the equation
0.80% of some mass (let's call it x) = 2.5 * 10^8 kg
Step 2: Solve for x
0.80% of x = 2.5 * 10^8 kg
To solve for x, we need to convert 0.80% to a decimal by dividing it by 100. So, 0.80% becomes 0.008.
0.008 * x = 2.5 * 10^8 kg
Now, divide both sides of the equation by 0.008 to solve for x:
x = (2.5 * 10^8 kg) / 0.008
Calculating this gives us the mass of ore needed, x.
Step 3: Convert mass to grams
Since the density of the ore is given in grams per cubic centimeter (g/cm^3), we need to convert the mass from kg to g.
1 kg = 1000 g
So, x grams = x * 1000 g
Step 4: Calculate volume
We know that mass = volume * density. Rearranging this equation, we get:
volume = mass / density
Now we have the mass in grams and the density in g/cm^3, so we can substitute these values to calculate the volume.
Let's say the density is 2.0 g/cm^3, so we have:
volume = (x * 1000 g) / 2.0 g/cm^3
Calculating this gives us the volume of ore removed each year.
To solve part (b), we need to calculate the mass of SO2 produced by roasting, assuming chalcopyrite is the only source of sulfur.
To do this, we will use stoichiometry.
Step 1: Write the equation and balance it
The roasting reaction of chalcopyrite can be represented as follows:
2CuFeS2 (s) + 3O2 (g) -> 2FeO (s) + 2CuS (s) + 2SO2 (g)
Step 2: Convert grams of Cu to moles
Use the molar mass of Cu to convert the mass of Cu to moles. The molar mass of Cu is approximately 63.55 g/mol.
Let's say the mass of Cu obtained is M. So, the moles of Cu can be calculated as:
moles of Cu = M / molar mass of Cu
Step 3: Convert moles of Cu to moles of SO2
From the balanced equation, we see that 1 mole of Cu reacts with 1 mole of SO2. So, the moles of SO2 can be calculated as:
moles of SO2 = moles of Cu
Step 4: Convert moles of SO2 to grams
Use the molar mass of SO2 to convert the moles of SO2 to grams. The molar mass of SO2 is approximately 64.06 g/mol.
mass of SO2 = moles of SO2 * molar mass of SO2
Calculating this gives us the mass of SO2 produced by roasting.
Remember, if you have any specific values for the mass of ore or other quantities, you can substitute them into the equations to calculate the final answers.