The balanced equation IS P4(s) + ioCl2 (g) 4PCl5 (g) late the et us er at Calcul

Question

The balanced equation IS P4(s) + ioCl2 (g) 4PCl5 (g) late the et us er at Calculations involving a limiting reactant Now consider a situation in which 27.0g of P, is added to 54.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations: 1. Calculate the number of moles of PCl5 that can be produced from 27.0g of P, (and excess Cla). 2. Calculate the number of moles of PCl5 that can be produced from 54.0 g of Cl2 (and excess P4). Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant. e a total s. The of ed Part B How many moles of PCl5 can be produced from 27.0 g of P, (and excess Cl.)? Express your answer to three significant figures and include the appropriate units. » Hints he slices ave It have pad is actant P: wÅ - ? | Value Units bdient, ke into Submit My Answers Give Up on the you 5. You Incorrect; Try Again y and

Explanation / Answer

Molar mass of P4 = 123.88 g/mol

mass of P4 = 27 g

mol of P4 = (mass)/(molar mass)

= 27/123.88

= 0.218 mol

From balanced chemical reaction, we see that

when 1 mol of P4 reacts, 4 mol of PCl5 is formed

mol of PCl5 formed = (4/1)* moles of P4

= (4/1)*0.218

= 0.8718 mol

Molar mass of PCl5 = 1*MM(P) + 5*MM(Cl)

= 1*30.97 + 5*35.45

= 208.22 g/mol

mass of PCl5 = number of mol * molar mass

= 0.8718*208.22

= 182 g

Answer: 182 g

Feel free to comment below if you have any doubts or if this answer do not work

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