Question

If 16.4 grams of copper (II) bromide react with 22.7 grams of sodium chloride, how many grams of sodium bromide are formed?

Answer 1

The amount of sodium bromide that would be formed from the reaction will be 7.5524 grams

Stoichiometric calculation

Looking at the equation of the reaction:

$CuBr_2 + 2NaCl --- > CuCl_2 + 2NaBr$

The mole ratio of CuBr2 and NaCl is 1:2.

Mole of 16.4 grams of CuBr2 = 16.4/223.37

                                                 = 0.0734 moles

Mole of 22.7 grams of NaCl = 22.7/58.44

                                                 = 0.3884 moles

Equivalent mole of NaCl = 0.1468 moles

Thus, NaCl is in excess while CuBr2 is limiting.

Mole ratio of CuBr2 and NaBr = 1:1

Mass of 0.0734 mole NaBr = 0.0734 x 102.894

                                              = 7.5524 grams

More on stoichiometric calculation can be found here: brainly.com/question/8062886