An emergency breathing apparatus often placed in mines works via the following chemical reaction: 4KO2(s) + 2CO2(g) to make 2K2CO3(s) + O2(g)
If the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breath while exiting the mine. Notice that the reaction produces O2, with can be breathed, and absorbs CO2, a product of respiration. What minimum amount of KO2 is required for the apparatus to produce enough oxygen to allow the user 15 minutes to exit in an emergency? Assume that an adult consumes approximately 4.4g of oxygen in 15 minutes of normal breathing.
remember question?
you said to find the moles of O2 which I think is 1.1 mol O2
but I got confused on the second part.
"it takes four times that amount of moles of O2 of KO2. What is the mass of that?" do I find the moles of that and then that is it?
I just noticed you equation is not balanced. The O on the left don't match the right.
4KO2 + 2CO2 >> 2K2CO3 + 3O2
Does that balance?
Now, if you want say 4.4 grams of O2, convert that to moles, then you need 4/3 of that much (4/3) moles of potassium superoxide KO2.
amountKO2= gramsO2/molmassO2 * 4/3*molmassKO2.
or 4.4g/32g * 4/3 * (40 + 32) check that mole mass of KO2.
= 4.4*4*72/32 grams KO2
The entire point of this is that the balanced equation is the guide in the mole ratios of the reactants and products.
Well, it seems like you've got the correct balanced equation now! Good job catching that. Now, let's go through the calculation.
First, convert the 4.4 grams of O2 to moles:
moles of O2 = 4.4g / molar mass of O2
Next, we need to find the moles of KO2 required. According to the balanced equation, for every 3 moles of O2 produced, we need 4 moles of KO2. So, the moles of KO2 required is:
moles of KO2 = (moles of O2) * (4/3)
Finally, we can find the mass of KO2:
mass of KO2 = (moles of KO2) * molar mass of KO2
Remember to use the correct molar mass for each compound.
Yes, you are correct. The balanced equation is:
4KO2 + 2CO2 >> 2K2CO3 + 3O2
To calculate the minimum amount of KO2 required for the apparatus to produce enough oxygen for the user to exit in 15 minutes, we can follow these steps:
1. Calculate the moles of oxygen needed for 15 minutes of breathing:
Given: 15 minutes = 4.4 grams of O2
Molar mass of O2 = 32 g/mol
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 4.4 g / 32 g/mol
moles of O2 = 0.1375 mol O2
2. Use the balanced equation to determine the mole ratio between O2 and KO2:
From the balanced equation, we can see that the mole ratio between O2 and KO2 is 3:4. This means that for every 3 moles of O2 produced, we need 4 moles of KO2.
3. Calculate the moles of KO2 needed:
moles of KO2 needed = 4/3 * moles of O2
moles of KO2 needed = 4/3 * 0.1375 mol
moles of KO2 needed = 0.1833 mol KO2
4. Convert moles of KO2 to grams:
Molar mass of KO2 = 39.1 g/mol + 16.0 g/mol + 16.0 g/mol = 71.1 g/mol
mass of KO2 needed = moles of KO2 needed * molar mass of KO2
mass of KO2 needed = 0.1833 mol * 71.1 g/mol
mass of KO2 needed = 13.03 g KO2
Therefore, the minimum amount of KO2 required for the apparatus to produce enough oxygen to allow the user 15 minutes to exit in an emergency is approximately 13.03 grams of KO2.
Yes, you are correct. The balanced equation for the chemical reaction is:
4KO2 + 2CO2 → 2K2CO3 + 3O2
To find the minimum amount of KO2 required to produce enough oxygen for the worker to exit the mine in 15 minutes, we need to calculate the amount of oxygen needed and then determine the corresponding amount of KO2.
Given that an adult consumes approximately 4.4 grams of oxygen in 15 minutes of normal breathing, we can first convert this mass of oxygen to moles:
moles of O2 = (mass of O2) / (molar mass of O2)
moles of O2 = 4.4 g / 32 g/mol (molar mass of O2)
moles of O2 = 0.1375 mol
According to the balanced equation, the ratio between the moles of O2 and KO2 is 3:4. So, we need 4/3 times the moles of O2 of KO2 to produce enough O2 for the worker to exit:
moles of KO2 = (moles of O2) * (4/3)
moles of KO2 = 0.1375 mol * (4/3)
moles of KO2 = 0.1833 mol
Finally, we can calculate the mass of KO2 needed using the molar mass of KO2:
mass of KO2 = (moles of KO2) * (molar mass of KO2)
mass of KO2 = 0.1833 mol * (40 + 16 + 16) g/mol (molar mass of KO2)
mass of KO2 = 4.996 g
Therefore, the minimum amount of KO2 required for the apparatus to produce enough oxygen to allow the user 15 minutes to exit in an emergency is approximately 4.996 grams.