cellowithgills Posted February 12, 2012 Share Posted February 12, 2012 I'm stuck in my chem homework. Here's the problem: Phosphorus can be prepared from calcium phosphate by the following reaction: 2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) ---> 6CaSiO3(s) + P4(s) + 10CO(g) Phosphorite is a mineral the contains Ca3(PO4)2 plus other non-phosphorus-containing compounds. What is the maximum amount of P4 that can be produced from 1.0kg of phosphorite if the phosephorite sample is 75% Ca3(PO4)2 by mass? Assume an excess of other reactants. Link to comment Share on other sites More sharing options...
reeferscooter Posted February 12, 2012 Share Posted February 12, 2012 I am not a chemistry type but i am a physics guy and i would do 75% of the 1kg mass then multiply that by the ratio between reactant and product to get a final mass. so basically .75kg times 1/2. Link to comment Share on other sites More sharing options...
markvo Posted February 12, 2012 Share Posted February 12, 2012 I'll give it a guess. 1 kg of sample, 75 % pure, atomic weight of P=15, Ca=20, and O=8, 4 molecules of P per P4 1000g * .75 * (15 * 2)/(20 * 3 + 8 * 8 + 15 * 2) * 1/4 = 36.53g Link to comment Share on other sites More sharing options...
Trautman Posted February 12, 2012 Share Posted February 12, 2012 This is a classic limiting reactant problem. Step 1: convert 1 kg (or really .75 x 1kg =.75 kg) Phosphorite to moles which gives you X moles of P Step 2: The Multiply by the coefficient of where you are coming from divided by where you are going to... so basically do nothing because it would be 4/4 or 1 So basically, my answer is: whatever you get for step one. Now I could be wrong, so check out this website which will actually tell you the right way to do it: khanacademy.com Link to comment Share on other sites More sharing options...
Trautman Posted February 12, 2012 Share Posted February 12, 2012 http://www.khanacademy.org/video/limiting-reactant-example-problem-1?topic=chemistry Link to comment Share on other sites More sharing options...
mrwheeling Posted February 12, 2012 Share Posted February 12, 2012 Start with the mass of your phosphorite sample (75% of 1.0kg which would be .75kg). Next Calculate your molar mass (markvo showed you how to use the atomic weights of each element and multiply each by how many atoms you have). Next use your mole ratio, 2:1 (2 molecules of calcium phosphate to one molecule of P4). Then calculate your molar mass of P4. Then solve. (I'm hoping you know how to use a conversion table) mass of sample X (1mole/molar mass of calcium phosphate sample) X (1 mole P4/2 moles of phosphate) X (molar mass of P4/1 mole) That is probably confusing, if you're still stuck tomorrow I'd be glad to help you. I get .15kg, I'll check that in the morning when I'm awake. Link to comment Share on other sites More sharing options...
Burningbaal Posted February 12, 2012 Share Posted February 12, 2012 Start with the mass of your phosphorite sample (75% of 1.0kg which would be .75kg). Next Calculate your molar mass (markvo showed you how to use the atomic weights of each element and multiply each by how many atoms you have). Next use your mole ratio, 2:1 (2 molecules of calcium phosphate to one molecule of P4). Then calculate your molar mass of P4. Then solve. (I'm hoping you know how to use a conversion table) mass of sample X (1mole/molar mass of calcium phosphate sample) X (1 mole P4/2 moles of phosphate) X (molar mass of P4/1 mole) That is probably confusing, if you're still stuck tomorrow I'd be glad to help you. I get .15kg, I'll check that in the morning when I'm awake. I didn't actually do the math, but this is the right system. Link to comment Share on other sites More sharing options...
ThePremiumAquarium Posted February 12, 2012 Share Posted February 12, 2012 ewww chemistry... I believe that mrwheeling has it right from what I can remember(been too long and thanks for the refresher). I didn't do the math either though so double check the answer. Link to comment Share on other sites More sharing options...
cellowithgills Posted February 12, 2012 Author Share Posted February 12, 2012 I ended up with 150g. I worked it out and then the next 5 problems did the same thing and drilled it into my head. Link to comment Share on other sites More sharing options...
markvo Posted February 12, 2012 Share Posted February 12, 2012 I see that I made a mistake with the final multiplying by 1/4. Removing that final step gives me the following: 1000g * .75 * (15 * 2)/(20 * 3 + 8 * 8 + 15 * 2) = 146g Link to comment Share on other sites More sharing options...
dsoz Posted February 12, 2012 Share Posted February 12, 2012 I'll give it a guess. 1 kg of sample, 75 % pure, atomic weight of P=15, Ca=20, and O=8, 4 molecules of P per P4 1000g * .75 * (15 * 2)/(20 * 3 + 8 * 8 + 15 * 2) * 1/4 = 36.53g You are using the atomic number, not the atomic mass. Ca=40.08, P=30.97, O=15.9994. 750g Ca3(PO4)2 * (1mol Ca3(PO4)2 / 310.24 g Ca3(PO4)2) * ( 1 mol P4 / 2 mol Ca3(PO4)2 ) * (123.88 g P4 / 1 mol P4) = 149.7 g P4 Make it 150 g P4 (significant figures) and it is all good. Link to comment Share on other sites More sharing options...
MVPaquatics Posted February 12, 2012 Share Posted February 12, 2012 Oh no. Not significant figures! I was just gonna add they used number not mass, beat me to it Link to comment Share on other sites More sharing options...
markvo Posted February 12, 2012 Share Posted February 12, 2012 Atomic mass is always twice the atomic number within reason. The electrons weigh so little compared to the protons and neutrons that they can be ignored for such large number of molecules as we deal with in normal situations. Since we only need to get the weight of the P relative to the compound, we can use the atomic weight of the elements. Link to comment Share on other sites More sharing options...
C0lin Posted February 12, 2012 Share Posted February 12, 2012 You know, there aren't a lot of places I go where I can feel dumb, but this is one of them. Kudos! Link to comment Share on other sites More sharing options...
dsoz Posted February 12, 2012 Share Posted February 12, 2012 Atomic mass is always twice the atomic number within reason. The electrons weigh so little compared to the protons and neutrons that they can be ignored for such large number of molecules as we deal with in normal situations. Since we only need to get the weight of the P relative to the compound' date=' we can use the atomic weight of the elements.[/quote'] Not to be negative, but this is only true up to about element 26 (iron). After that the pattern breaks down. Some notable elements that I can think of off the top of my head... gold #79 with a mass of 197. uranium #92 mass of 238. mercury #80 mass of 200. Even Zinc #30 with a mass of 65... After iron, there needs to be a greater than 1:1 ratio of protons:neutrons to make the element stable. For all practical purposes, light elements (like those in this problem) your premise of mass=2(number) holds fairly true. But don't extrapolate the answer further than the limits of small, light, elements. In context, of the original post (being stuck with homework), I feel it is better to give a more complete answer so the process can be repeated. If you give a partial answer that is not true in some situations, you are setting up cellowithgills to not be consistent with producing correct answers. This is done best by stoicheometry (converting between grams and moles of the two substances in question). Sorry, it is just the chemistry teacher in me that strives for better understanding. dsoz Link to comment Share on other sites More sharing options...
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