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The limiting reactant when reacting 88g of Sodium with 88g of Chlorine to get Sodium Chloride is the?

The limiting reactant when reacting 88g of Sodium with 88g of Chlorine to get Sodium Chloride is the?

Grade:12

2 Answers

Vikas TU
14149 Points
9 years ago
Before determining the limiting reactant, one must take into account the mole ratio. Therefore, always start with a balanced chemical equation. 

2Na(s) + Cl2(g) --> 2NaCl(s) 
88g ........88g ...........?g 

If you have a question about the amount of product, then you can use the amounts of each reactant to compute the amount of product. The lesser amount of product is the theoretical yield, and the reactant which gives the smaller amount of product is the limiting reactant. 
 
Vikas TU
14149 Points
9 years ago
Or you can pick one reactant as the "limiting reactant" and see how much of the other reactant you will need. Suppose we assume Na is the limiting reactant. How much Cl2 will be needed? 

88g Na x (1 mol Na / 23.0g Na) x (1 mol Cl2 / 2 mol Na) x (70.90g Cl2 / 1 mol Cl2) = 135.6g CL2 

88 g Na requires 135.6g Cl2, but only 88g Cl2 is available. Therefore, Cl2 is the limiting reactant.

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