Your starting point here will be the balanced chemical equation for this synthesis reaction. Aluminium metal, #"Al"#, will react with chlorine gas, #"Cl"_2#, to produce aluminium chloride, #"AlCl"_3#, according to the balanced chemical equation
Now, you know that aluminium is in excess, which implies that the reaction will completely consume the chlorine gas. Notice that you have a #color(red)(3):color(blue)(2)# mole ratio between chlorine gas and aluminium chloride. This tells you that for every three moles of chlorine gas that take part in the reaction, you get two moles of aluminium chloride. Use the molar mass of chlorine gas to determine how many moles you have in that #"80.0-g"# sample
Next, use the aforementioned mole ratio to determine how many moles of aluminium chloride will be produced by this many moles of chlorine gas
Finally, to find the mass of aluminium chloride that would contain that many moles, use the compound's molar mass
Rounded to three **sig figs, the number of sig figs you have for the mass of chlorine gas, the answer will be
Melanie L. asked • 04/14/21
Use the equation 2 Al + 3 Cl2 —> 2 AlCl3. If 2 moles of aluminum and 2 moles of chlorine are reacted, identify the limiting reactant 1 Expert Answer Hi, Melanie, The balanced equation 2 Al + 3 Cl2 —> 2 AlCl3 tells us that it takes 3 moles of Cl2 to completely react with 2 moles of Al. If they aren't in the ratio of 3 moles Cl2 to 2 moles Al, there will be a surplus of one relative to the other. The one in shortest supply is termed the "limiting reagent." It means that there is not enough of it to completely consume the other reactant(s). If 2 moles of aluminum and 2 moles of chlorine are reacted, we can use the molar ratio of (2 moles Al/3 moles Cl2) to tell us how much Al we would need for the 2 moles Cl2: (2 moles Al/3 moles Cl2)*(2 moles Cl2) = 4/3 moles Al. We have two moles, so there is plenty. The Cl2 must be the limiting reagent. We can also write the factor as (3 moles Cl2/2 moles Al), and use it to determine how much chlorine will be needed: (3 moles Cl2/2 moles Al)*(2 moles Al) = 3 moles Cl2 We only have 2 moles Cl2, so it is the limiting reagent. I hope this helps, Bob
In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. Q140029 Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction. 2Al (s) + 3Cl2 (g) --> 2AlCl3 (s) You are given 34.0g of aluminum and 39.0g of chlorine gas. 1-If you had excess chlorine, how many moles of aluminum chloride could be produced from 34.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. 2-If you had excess aluminum, how many moles of aluminum chloride could be produced from 39.0 g of chlorine gas, Cl2? Express your answer to three significant figures and include the appropriate units. Solution : Part 1) The given reaction is 2Al (s) + 3Cl2 (g) --> 2AlCl3 (s) mass of aluminum we are given = 34.0g , and Cl2 is in excess. So amount of aluminum chloride produced will depend on the mass of aluminum. Atomic mass of Al = 26.982g/mol moles of Al = 34.0g Al * 1mol Al / 26.982g Al = 1.2601 mol of Al Next we use the mole ratio of Al and AlCl3 from the reaction and find the ‘moles of AlCl3 ‘ . mol ratio of Al and AlCl3 in the reaction = 2 : 2 = 1 : 1 So, moles of AlCl3 formed = 1.2601 mol of Al * 1 mol AlCl3 /1 mol Al = 1.2601mol of AlCl3 In question we are told to write the answer in 3 significant figure. So, moles of Aluminum chloride formed from given mass of Aluminum = 1.26 mol of AlCl3 Part 2) In second part, we are given mass of chlorine gas, Cl2 = 39.0g and Aluminum is in excess. We will convert 39.0g of Cl2 to moles using molar mass of Cl2 . Atomic mass of Cl = 35.453g/mol Molar mass of Cl2 = 2 * atomic mass of Cl = 2 * 35.453g/mol = 70.906g/mol moles of Cl2 = 39.0g Cl2 * 1 mol Cl2 /70.906 g of Cl2 = 0.5500 mol of Cl2 . Next we will use the mol ratio of Cl2 and AlCl3 from the reaction and find the ‘moles of AlCl3 ‘ formed from 0.5500 mol of Cl2 2Al (s) + 3Cl2 (g) --> 2AlCl3 (s) mol ratio of Cl2 and AlCl3 in the given reaction is 3 : 2 moles of AlCl3 formed = 0.5500 mol Cl2 * 2 mol AlCl3 /3 mol Cl2 = 0.3667 mol AlCl3 which in 3 significant figure = 0.367 mol AlCl3 . Hence, moles of Aluminum chloride formed from given mass of chlorine gas is 0.367mol
1 Answers
The equation is 2Al+3Cl2=2AlCl3. N(AlCl3)=n(Al)=4 moles, so 4 moles of AlCl3 are produced. |