Question

lete reaction between 0.48 mol Na and 0.36 mol Cl2 for the formation of sodium chloride.

2 Na(s) + Cl2(g) ? 2 NaCl(s)

Answer 1

2Na(s) + Cl2(g) → 2NaCl(s) Initially mixed 0.48 mol 0.36 mol 0.00 mol How much reacts /is produced 0.24 mol 0.48 Final mixture mol 0.12

solution:

2 Na +   Cl2 ------------------------------------> 2 NaCl

2 mol     1 mol                                               2 mol

0.48       0.36

Na is limiting reagent . so total Na consumed .

2 mol Na --------------------> 2 mol NaCl

0.48 mol Na ---------------> 0.48 mol NaCl

2 mol Na ----------------> 1 mol Cl2

0.48 mol Na -------------> 0.48 / 2 = 0.24 mol Cl2 needed

remaining Cl2 = 0.36 - 0.24 = 0.12 mol