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a fost răspuns

calculati masa precipitatului format in urma interactiunii cu 200 g solutie de hidroxid de sodiu cu partea de masa 4% (NaOH)si o solutie de clorura de cupru (CuCl2)

Răspuns :

Pokju8888

          8g                                        xg
         2NaOH +CuCl2 =2 NaCl +Cu(OH)2                                                                                  2.40g                               98g
MNaOH=23+16+1=40-------->1mol=40g                                                                         MCu(OH)2 =64+2.16 +2=98-------->1mol=98g
Se calculeaza md de NaOH  din solutie                                                                     c=md.100:ms                                                                                                                  md=  c.ms:100                                                                                                                                                                                                                                                                                md=4.200:100=8g Na                                                                                                      Se calc. x de pe ecuatie                                                                                                                                           x=8.98:80=9,8 g Cu(OH)2

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