Show work please!
Consider the following reaction: P4O10 + 6 PCl5---> 10 POCl3
0.450 g of P4O10 is mixed with 1.5000 g of PCl5 and allowed to react as shown above. What is the mass of each substance present In the equation when the reaction has come to completion?
A) g of P4O10
B) g of PCl5
C) g of POCl3
Need help with chemistry calculations. Find the mass of each substance!?
2015-05-28 6:41 pm
回答 (1)
2015-05-28 9:00 pm
(0.450 g P4O10) / (283.8890 g P4O10/mol) = 0.0015851 mol P4O10
(1.5000 g PCl5) / (208.2388 g PCl5/mol) = 0.007203269 mol PCl5
0.007203269 mole of PCl5 would react completely with 0.007203269 x (1/6) = 0.0012005 mole of P4O10, but there is more P4O10 present than that, so P4O10 is in excess and PCl5 is the limiting reactant.
A)
((0.0015851 mol P4O10 initially) - (0.0012005 mol P4O10 reacted)) x (283.8890 g P4O10/mol) =
0.109 g P4O10 left over
B) There is no PCl5 present when the reaction reaches completion because PCl5 is the limiting reactant, and so was completely consumed.
C)
(0.007203269 mol PCl5) x (10 mol POCl3 / 6 mol PCl5) x (153.3322 g POCl3/mol) = 1.841 g POCl3
[Checking:
0.450 g + 1.5000 g = 1.95 g
0.109 g + 1.841 g = 1.95 g ]
(1.5000 g PCl5) / (208.2388 g PCl5/mol) = 0.007203269 mol PCl5
0.007203269 mole of PCl5 would react completely with 0.007203269 x (1/6) = 0.0012005 mole of P4O10, but there is more P4O10 present than that, so P4O10 is in excess and PCl5 is the limiting reactant.
A)
((0.0015851 mol P4O10 initially) - (0.0012005 mol P4O10 reacted)) x (283.8890 g P4O10/mol) =
0.109 g P4O10 left over
B) There is no PCl5 present when the reaction reaches completion because PCl5 is the limiting reactant, and so was completely consumed.
C)
(0.007203269 mol PCl5) x (10 mol POCl3 / 6 mol PCl5) x (153.3322 g POCl3/mol) = 1.841 g POCl3
[Checking:
0.450 g + 1.5000 g = 1.95 g
0.109 g + 1.841 g = 1.95 g ]
收錄日期: 2021-05-01 20:43:11
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