Question

A reaction is carried out where 8.00 g of sodium metal were reacted with chlorine gas, producing 20.0 g of NaCI. 2 Na (s) + CI_2 (g) rightarrow 2 NaCI (s) How many moles of NaCI were produced? How many mols of Cl_2 must have reacted? How many grams is this?

Answer 1

1)molecular weight of NaCl = 58.44 g/mol

58.44 = 1.0 mole

20.0 g = 20 / 58.44 =0 .34 moles

so 0.34 moles NaCl produced.

2) cording to balaced reaction

2 moles of NaCl produced by consuming 1 mole Cl2

0.34 moles NaCl produced by consuming (0.34 x 1.0) / 2= 0.17 moles Cl2

1 mole Cl2 = 71 g

0.17 moles = 0.17 x 71 = 12.07g Cl2