Hi!! Can someone please layout the steps on how to solve M11...I am having trouble figuring out which equation to pull from for coefficients in order to solve for the limiting reactants.
Here is the problem: A reaction vessel contains 5.77 g of white phosphorus and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorous (III) oxide, P406; P4 (s) + 3O2 (g) --> P406(s). IF enough oxygen is present, the oxygen can react further w/ this oxide to produce phosphorus (V0 oxide, P4010; P406 (s) + 2 O2 (g) --> P4010 (s). (a) What is the limiting reactant for the formation of P4010? (b) What mass of P4010 is produced? (c) how many grams of the excess reactant remain in the reaction vessel?
Fundamentals M 11
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Kristy Sun 3L
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Re: Fundamentals M 11
Hello!
(a) First, you would to find the limiting reactant, you can start by finding how many moles 5.77g of P4 is equal to. So, by dividing 5.77 g by 123.892 g/mol, you would get 0.0466 mol of P4. Since the ratio of P4 to O2 is 1 to 3 in the first reaction, you would multiply 0.0466 mol by 3 to find how many moles of Oxygen is needed in the first reaction, which is 0.1398 mol in this case. Then, to find how many grams of O2 it uses, you would multiply 0.1398 mol by the molar mass of O2, which is about 32 g/mol, to get 4.476 grams of O2. You are told in the question that you have 5.77 g of Oxygen total, so 5.77 g - 4.476 g = 1.3 g of O2 left to be used.
Then, we take a look at the second reaction. The second reaction uses the product of the first reaction and oxygen to produce P4O10. In the steps above, you calculated that 0.0466 mol of P4 is being used in the reaction. Since, in reaction 1, the ratio of P4 to P4O6 is 1:1, we know that 0.0466 mol of P4O6 would be produced. Therefore, for the second reaction, there will be 0.0466 mol of P4O6. As for oxygen, we calculated above that there is 1.3 g of O2 left to be used for the second reaction. To find the equivalent amount of moles, we divide 1.3 g by 32 g/mol to find 0.0406 mol. The ratio of P4O6 to O2 is 1:2 in the second reaction, so 0.0406 mol of O2 would mean that 0.0203 mol of P4O6 is being used up in the reaction. However, we have a total of 0.0466 mol of P4O6. Since we are not using up all of P4O6, this would mean that O2 is a limiting reactant.
(b) In the steps above, we found that, in reaction 2, 0.0203 mol of P4O6 will be reacting with 0.0406 mol of O2. Since the ratio of P4O6 and P4O10 is 1:1, we know that 0.0203 mol of P4O10 is being produced. Therefore, to find the mass of P4O10, you would multiply 0.0203 mol by the molar mass of P4O10, which is 283.886 g/mol in this case. Therefore, the answer would be 5.763 g of P4O10.
(c) We know from the calculations that the limiting reactant is O2, which means that the excess reactant is P4O6. To find how many grams of the excess reactant remain in the reaction vessel, you would first find how many grams of this reactant was produced in the first place. In the first reaction, 0.0466 mol of P4O6 was produced. To find how many grams of P4O6 was produced, we multiply 0.0466 mol by 219.8884 g/mol to get 10.246 g of P4O6. That is the total amount of P4O6. In the second reaction, we used up 0.0203 mol of P4O6. To find how many grams this is equivalent to, we multiply 0.0203 by 219.8884 g/mol to get 4.47 g of P4O6 used in the second reaction. Therefore, we would subtract 10.246 g - 4.47 g to get 5.78 g of P4O6 that remained in the reaction vessel.
Hope this helps!
(a) First, you would to find the limiting reactant, you can start by finding how many moles 5.77g of P4 is equal to. So, by dividing 5.77 g by 123.892 g/mol, you would get 0.0466 mol of P4. Since the ratio of P4 to O2 is 1 to 3 in the first reaction, you would multiply 0.0466 mol by 3 to find how many moles of Oxygen is needed in the first reaction, which is 0.1398 mol in this case. Then, to find how many grams of O2 it uses, you would multiply 0.1398 mol by the molar mass of O2, which is about 32 g/mol, to get 4.476 grams of O2. You are told in the question that you have 5.77 g of Oxygen total, so 5.77 g - 4.476 g = 1.3 g of O2 left to be used.
Then, we take a look at the second reaction. The second reaction uses the product of the first reaction and oxygen to produce P4O10. In the steps above, you calculated that 0.0466 mol of P4 is being used in the reaction. Since, in reaction 1, the ratio of P4 to P4O6 is 1:1, we know that 0.0466 mol of P4O6 would be produced. Therefore, for the second reaction, there will be 0.0466 mol of P4O6. As for oxygen, we calculated above that there is 1.3 g of O2 left to be used for the second reaction. To find the equivalent amount of moles, we divide 1.3 g by 32 g/mol to find 0.0406 mol. The ratio of P4O6 to O2 is 1:2 in the second reaction, so 0.0406 mol of O2 would mean that 0.0203 mol of P4O6 is being used up in the reaction. However, we have a total of 0.0466 mol of P4O6. Since we are not using up all of P4O6, this would mean that O2 is a limiting reactant.
(b) In the steps above, we found that, in reaction 2, 0.0203 mol of P4O6 will be reacting with 0.0406 mol of O2. Since the ratio of P4O6 and P4O10 is 1:1, we know that 0.0203 mol of P4O10 is being produced. Therefore, to find the mass of P4O10, you would multiply 0.0203 mol by the molar mass of P4O10, which is 283.886 g/mol in this case. Therefore, the answer would be 5.763 g of P4O10.
(c) We know from the calculations that the limiting reactant is O2, which means that the excess reactant is P4O6. To find how many grams of the excess reactant remain in the reaction vessel, you would first find how many grams of this reactant was produced in the first place. In the first reaction, 0.0466 mol of P4O6 was produced. To find how many grams of P4O6 was produced, we multiply 0.0466 mol by 219.8884 g/mol to get 10.246 g of P4O6. That is the total amount of P4O6. In the second reaction, we used up 0.0203 mol of P4O6. To find how many grams this is equivalent to, we multiply 0.0203 by 219.8884 g/mol to get 4.47 g of P4O6 used in the second reaction. Therefore, we would subtract 10.246 g - 4.47 g to get 5.78 g of P4O6 that remained in the reaction vessel.
Hope this helps!
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