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

If 164 Grams Of Copper II Bromide React With 227 Grams Of Sodium Chloride How Many Grams Of Sodium Bromide Are Formed class=

Sagot :

Oseni

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:

[tex]CuBr_2 + 2NaCl --- > CuCl_2 + 2NaBr[/tex]

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: https://brainly.com/question/8062886

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