Aluminum reacts with chlorine gas to form aluminum chloride via the following re

Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)2AlCl3(s) You are given 31.0 g of aluminum and 36.0 g of chlorine gas. Part A If you had excess chlorine, how many moles of of aluminum chloride could be produced from 31.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. 1.15 mol SubmitHintsMy AnswersGive UpReview Part Correct Part B If you had excess aluminum, how many moles of aluminum chloride could be produced from 36.0 g of chlorine gas, Cl2? Express your answer to three significant figures and include the appropriate units. 0.338 mol SubmitHintsMy AnswersGive UpReview Part Correct By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces the lesser amount of product. Significant Figures Feedback: Your answer 0.339mol was either rounded differently or used a different number of significant figures than required for this part. Part C Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 31.0 g of aluminum with 36.0 g of chlorine? Express your answer to three significant figures and incl

Explanation / Answer

A ) No. of moles of Al = 31 / 26.98 = 1.14

One mole of Al is reacted,

Therefore 1.14 moles of AlCL3 are produced.

B ) No of moles of Cl = 36 / 35.5 x 3 = 0.338 moles ( In AlCl3 there are 3 Cl atoms therefore multiplied by 3)

Aluminium is limiting reactants.

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