Rosalina Posted May 12, 2016 Report Share Posted May 12, 2016 Hi there, I was wondering if anyone could help me with this question since I have no answer key whatsoever. Identify the limiting and excess reagents for each of the following pairs of reactants, and determine the chemical amount in excess in each case. a) Cl2 (aq) + NaI (aq) -----> 10 mol 10 mol b) AlCl3(aq) + NaOH(aq) -----> 20 g 20 g I was able to determine limiting reagents in a and b ( NaI and AlCl3) but am not sure what the chemical amount is or how to get it. Some clarification would be great Thanks in advance! Reply Link to post Share on other sites More sharing options...
Msj Chem Posted May 12, 2016 Report Share Posted May 12, 2016 (edited) To calculate the excess reactant remaining, you need to find the molar ratio of reactants, and then subtract the amount of the limiting reactant that has reacted from the amount of the excess. This will give you the amount of excess reactant remaining. For example: 2NaI + Cl2 ==> 2NaCl + I2 NaI 10/2 = 5 Cl2 10/1 = 10 NaI is limiting reactant 10 mol of NaI reacts with 5 mol of Cl2 (2:1 ratio). mol of Cl2 remaining 10 - 5 = 5 mol Edited May 12, 2016 by Msj Chem 1 Reply Link to post Share on other sites More sharing options...
Rosalina Posted May 12, 2016 Author Report Share Posted May 12, 2016 17 minutes ago, Msj Chem said: To calculate the excess reactant remaining, you need to find the molar ratio of reactants, and then subtract the amount of the limiting reactant that has reacted from the amount of the excess. This will give you the amount of excess reactant remaining. For example: 2NaI + Cl2 ==> 2NaCl + I2 NaI 10/2 = 5 Cl2 10/1 = 10 NaI is limiting reactant 10 mol of NaI reacts with 5 mol of Cl2 (2:1 ratio). mol of Cl2 remaining 10 - 5 = 5 mol Just to clarify, why shouldn't the Cl2 be like this: Cl2 : (10 mol) (2 mol NaI/1 mol Cl2) Reply Link to post Share on other sites More sharing options...
Msj Chem Posted May 12, 2016 Report Share Posted May 12, 2016 (edited) If you want to use dimensional analysis, then: 10 mol Cl2 x (2 mol NaI / 1 mol Cl2) = 20 mol NaI This tells you that 10 mol of Cl2 would need 20 mol of NaI to react completely, but we only have 10 mol so NaI is the limiting reactant. I used this: 10 mol NaI x (1 mol Cl2 / 2 mol NaI) = 5 mol Cl2 This tells you that 10 mol NaI reacts with 5 mol Cl2. Edited May 12, 2016 by Msj Chem 1 Reply Link to post Share on other sites More sharing options...
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