13. Р-(s) P4(s) + _Cl2(g) → _PCI5(g) AH°, = -1500 kJ balance & classify: The reaction is _thermic, heat flows bonds between atoms become er In a reaction mixture of 0.22 g(P4) and 1.45 g(Cl2): significant figures! The 0.22 grams of P4 are enough to form mol(PCI5) LSTP (PCI5) The 1.45 grams of Cl2 are enough to form mol(PCI5) LSTP (PCI5) The limiting reagent (LR) in the mixture is The theoretical yield of PCI5 is mol and LSTP (liters at STP) If 0.12 LSTP(PCI5) was isolated, the percent reaction yield is % The most heat this reaction mixture can form is kJ

Chemistry for Engineering Students
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Author:Lawrence S. Brown, Tom Holme
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Chapter4: Stoichiometry
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Problem 4.72PAE: 4.72 The picture shown depicts the species present at the start of a combustion reaction between...
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P4(s) + _Cl2(g) → _PCI5(g)
AH°rxn = -1500 kJ
balance & classify:
-
--
The reaction is
_thermic, heat flows
bonds between atoms become
er
In a reaction mixture of 0.22 g(P4) and 1.45 g(Cl2):
significant figures!
The 0.22 grams of P4 are enough to form
mol(PCI5)
LSTP (PCI5)
The 1.45 grams of Cl, are enough to form
mol(PCI5)
LSTP (PCI5)
The limiting reagent (LR) in the mixture is
The theoretical yield of PCI5 is
mol and
LSTP (liters at STP)
If 0.12 LSTP(PCI5) was isolated, the percent reaction yield is
%
The most heat this reaction mixture can form is
kJ
Transcribed Image Text:P4(s) + _Cl2(g) → _PCI5(g) AH°rxn = -1500 kJ balance & classify: - -- The reaction is _thermic, heat flows bonds between atoms become er In a reaction mixture of 0.22 g(P4) and 1.45 g(Cl2): significant figures! The 0.22 grams of P4 are enough to form mol(PCI5) LSTP (PCI5) The 1.45 grams of Cl, are enough to form mol(PCI5) LSTP (PCI5) The limiting reagent (LR) in the mixture is The theoretical yield of PCI5 is mol and LSTP (liters at STP) If 0.12 LSTP(PCI5) was isolated, the percent reaction yield is % The most heat this reaction mixture can form is kJ
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