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What is limiting reactant in chemistry?

Help with chemistry limiting reactant problem. i did some of it but im not sure if its correct?

  • When 0.855g of Cl2 and 3,205g of KBr are mixed in a solution, which is the limiting reactant? And how many grams of KCl are produced?? 2KBr + Cl2 --> 2KCl + Br2 this is what i have so far: 0.855g Cl2 / 70.9g Cl2 = 0.012 mol Cl2 3.205g KBr / 119g KBr = 0.027 mol KBr so is that right? and is Cl2 the limiting reactant?? and most importantly how do you find out how much KCl is produced??!!

  • Answer:

    Yes your correct. Since the KBr and CL2 react in a 2:1 ratio, the actual amount of KBr needed is only 0.012x2 = 0.024. The simplest way to do the next part is to look at the stoch-ratios. 2KBr reacts with Cl2 to produce 2 moles of KCl and one mole of Br2. So if i had 2 moles of KBr i would get 2 moles of KCl and one mole of Br2. We can use this fact to find the next step. Try it yourself, the ratio for KBr to KCl is 1:1. (Remember not all the KBr reacts! Only use how much actually reacts!) My final answer is 1.79 grams of KCl produced.

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after finding the moles of each reactant, you would need to use mole relationships or stoichiometry to find the amount of KCl produced. 0.012 mol Cl2 (2 mole KCl / 1 mole Cl2) = 0.024 mole KCl 0.027 mol KBr (2 mol KCl / 2 mol KBr) = 0.027 mol KCl Since less KCl was produced from the Cl2 that means that Cl2 is the limiting reactant.

2KBr + Cl2 --> 2KCl + Br2 You are right to identify the limiting reactant as Cl2 Now from the above reaction 1 mole Cl2 reacting with 2 moles of KBr yields 2 moles KCl So 0.012 mole Cl2 will yield 2x0.012 = 0.024 moles KCl Thus the amount of KCl produced = 74.5x0.024 = 1.8 g

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