Question

Chemistry 1215, Experiment #10; The reaction of zinc and iodine, Pre-lab Name ____________________________________ Sodium metal reacts with chlorine gas according to the equation below. A student placed 8.20 g of sodium in a jar containing 20.62 g of Cl2. Answer the questions below concerning this reaction. 2Na(s) + Cl2(g) J 2NaCl(s) 1. How many moles of Na were originally present? 2. How many moles of chlorine were originally present? 3. Which reagent is limiting? 4. How many grams of product can be formed?

Answer 1

2Na(s) + Cl2(g) --------------> 2NaCl(s)

1) No. of moles of Na present = 8.20/ 23 = 0.3565 moles

2) No. of moles of Cl2 present = 20.62 / 70.9 = 0.2908 moles

3) From the equation; each mole of Cl2 needed 2 moles of Na

No. of moles of Na needed for 0.2908 moles of Cl2 = 2 x 0.2908 = 0.58166 moles

Actual Na is less than which is needed for Cl2. So, chlorine is excess

Limiting reagent is Na (One which is present in limited quantity).

4) 2 moles of Na will give 2 moles of NaCl

So, 0.3565 moles of Na will react with excess Cl2 will give NaCl = 0.3565 moles of NaCl

Weight of naCl produced = 0.3565 x 58.44 = 20.834 grams of NaCl formed