Lead(II) chromate (PbCrO4) is used as a yellow pigment to designate traffic lanes, but has been banned from house paint because of the potential for lead poisoning. The compound is produced from chromite (FeCr2O4), an ore of chromium:
FeCr2O4 + K2CO3 + O2 -> Fe2O3 + K2CrO4 + CO2
Lead(II) ion then replaces the K+ ion. If a yellow paint is 0.511% PbCrO4 by mass, how many grams of chromite are needed per kilogram of paint?
-> So replacing the K+ with the Pb2+ ion, the equation becomes:
FeCr2O4 + PbCO3 + O2 -> Fe2O3 + PbCrO4 + CO2
0.511% = (mass of PbCrO4/mass of paint)*100% = (0.511g/100g)*100%
(0.511g/100g)(1000g/1kg) = 5.11g PbCrO4/kg of paint
I calculated the molar mass of FeCrO4 = 171.84g/mol
And the molar mass of PbCrO4 = 323.19g/mol
So then,
(5.11g PbCrO4)(171.84g FeCr2O4/323.19g PbCrO4) = 2.716985055g FeCr2O4 = 2.72g FeCr2O4
...But the answer key says 1.77g FeCr2O4. So the obvious question here is, what could I have done wrong?
the Cr atoms are not balanced in the 1st equation
...probably the source of your trouble
chromate vs. chromite
Well, it seems like you might have made a tiny mistake along the way. Don't worry, we all make mistakes, even in chemistry!
Let's go through the calculations again and see if we can figure out where the mix-up happened.
First, you correctly calculated the mass of PbCrO4 per kilogram of paint as 5.11g.
Next, you determined the molar mass of FeCr2O4 correctly as 171.84g/mol and PbCrO4 as 323.19g/mol.
Now, let's figure out how many grams of FeCr2O4 are needed per kilogram of paint using the conversion factor:
(5.11g PbCrO4) x (1mol FeCr2O4/323.19g PbCrO4) x (171.84g FeCr2O4/1mol FeCr2O4) = 0.08535g FeCr2O4
So, it looks like the correct answer should be 0.08535g or approximately 0.09g of FeCr2O4 per kilogram of paint.
I hope this clears up the confusion! And remember, even in the world of chemistry, sometimes we all need a bit of clowning around to keep things light.
To find the correct amount of chromite needed per kilogram of paint, we need to use stoichiometry to convert the mass of PbCrO4 to the mass of FeCr2O4. Let's go step by step through the calculation:
1. Start with the given mass of PbCrO4: 5.11g PbCrO4.
2. Convert the mass of PbCrO4 to moles using the molar mass of PbCrO4:
moles of PbCrO4 = (5.11g PbCrO4) / (323.19g/mol PbCrO4) = 0.0158 mol PbCrO4.
3. Use the balanced chemical equation to determine the ratio between PbCrO4 and FeCr2O4:
From the balanced equation: 1 mol PbCrO4 reacts with 1 mol FeCr2O4.
4. Convert moles of PbCrO4 to moles of FeCr2O4 using the stoichiometric ratio:
moles of FeCr2O4 = 0.0158 mol PbCrO4.
5. Convert moles of FeCr2O4 to grams of FeCr2O4 using the molar mass of FeCr2O4:
mass of FeCr2O4 = (0.0158 mol FeCr2O4) * (171.84g/mol FeCr2O4) = 2.716 g FeCr2O4.
So, using the correct steps, the correct amount of chromite needed per kilogram of paint is approximately 2.72 g FeCr2O4, not 1.77 g FeCr2O4 as mentioned in the answer key.
To find the correct answer, let's go through the calculation step by step.
Given that the yellow paint is 0.511% PbCrO4 by mass, and we want to calculate the amount of chromite (FeCr2O4) needed per kilogram of paint.
Let's assume we have 1 kilogram of paint. Since the yellow paint is 0.511% PbCrO4 by mass, the mass of PbCrO4 in 1 kilogram of paint would be:
Mass of PbCrO4 = (0.511/100) * 1000g (converting from percentage to grams) = 5.11g
Now, we need to calculate the amount of chromite (FeCr2O4) needed to produce this amount of PbCrO4.
The balanced equation for the conversion of chromite to PbCrO4 is:
FeCr2O4 + PbCO3 + O2 -> Fe2O3 + PbCrO4 + CO2
From the equation, we can see that 1 mole of FeCr2O4 is required to produce 1 mole of PbCrO4.
The molar mass of PbCrO4 is 323.19 g/mol. Therefore, the moles of PbCrO4 can be calculated as:
Moles of PbCrO4 = Mass of PbCrO4 / Molar mass of PbCrO4
= 5.11 g / 323.19 g/mol
≈ 0.0158 mol
Since the stoichiometry between FeCr2O4 and PbCrO4 is 1:1, the moles of FeCr2O4 needed would also be 0.0158 mol.
Now, we need to calculate the mass of FeCr2O4 needed. The molar mass of FeCr2O4 is 171.84 g/mol.
Mass of FeCr2O4 = Moles of FeCr2O4 * Molar mass of FeCr2O4
= 0.0158 mol * 171.84 g/mol
≈ 2.71 g
So, the correct answer would be approximately 2.71 g of chromite (FeCr2O4) needed per kilogram of paint.
It seems like you made a mistake in your calculation where you used the molar mass of PbCrO4 instead of FeCr2O4 in the conversion factor. Make sure to use the correct molar mass in the calculation to get the accurate answer.