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P4(s)+Cl2(g)--PCl5(g) balance the chemical reaction equation?

source : yahoo.com

P4(s)+Cl2(g)–PCl5(g) balance the chemical reaction equation?

Enter the coefficients in order, separated by commas (e.g., 1,2,3).

#2

Consider a situation in which 186g of P4 are exposed to 208g of O2.

What is the maximum amount of P2O5 that can theoretically be made from 186g of P4 and excess oxygen?

What is the maximum amount of P2O5 that can theoretically be made from 208g of O2 and excess Phosphorus?

Thank you very much for your help. This chapter is hard!

Given the following reactions P 4 s 10 Cl 2 g 4 PCl... | Course Hero

Given the following reactions P 4 s 10 Cl 2 g 4 PCl… | Course Hero – This preview shows page 5 – 8 out of 8 pages. 15. Given the following reactions P4(s) + 10 Cl2(g) → 4 PCl5(s) ΔrH = -1774.0 kJ/mol PCl3(l) + Cl2(g) ___ 18. What is the correct standard thermochemical expression for the formation of H2O? a. H2(g) + ½ O2(g) → H2O(l) ΔH° = -285.8 kJ b. H2(g) + ½ O2…Phosphorus pentachloride is the chemical compound with the formula PCl5. It is one of the most important phosphorus chlorides, others being PCl3 and POCl3. PCl5 finds use as a chlorinating reagent.In this video we'll balance the equation P4 + Cl2 = PCl5 and provide the correct coefficients for each compound.To balance P4 + Cl2 = PCl5 you'll need to be…

Phosphorus pentachloride – Wikipedia – P4 + 6Cl2 ==> 4PCl3 balanced equation. (2) convert 12.3 g P4 to moles and calculate theoretical yield of PCl3 in moles (3) covert moles PCl3 to grams PCl3 theoretically formed: 0.3972 moles PCl3 x 137.3 g/mol = 54.5g PCl3 = theoretical yield.P4(s) + 6 Cl2(g)–> 4 PCl3(l) ΔH°f = ? The percentage yield for the reaction PCL3+Cl2–>PCL5 is 83.2%. What mass of PCL5 is expected from the reaction of 73.7g of PCL3 with excess chlorine?What is the maximum amount of P2O5 that can theoretically be made from 186g of P4 and excess oxygen? P4(s) + 10 Cl2(g) → 4 PCl5(g).

Phosphorus pentachloride - Wikipedia

How to Balance P4 + Cl2 = PCl5 – YouTube – Calculate the number of moles of PCl5 that can be produced from 53.0 g of Cl2 (and excess P4). Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant.4PCl₅(g)⇌P₄(s)+10Cl₂(g) Kgoal = ?Determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=? by making use of the following information: P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019 PCl5(g)⇌PCl3(g)+Cl2(g)

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