Question

determinati partea de masa a sarii obtinute in urma reactiei a 40 cm cub de hidroxid de potasiu de 6,2%(densitatea =1,055 g\cm cub)cu 10 cm cub de acid fosforic(densitatea =1,075 g\cm cub) cu partea de masa 13,76%

Answer 1

din densitate ==>ro=ms/Vs==>ms=ro*Vs=1.055*40=42.2g
Cp=md/ms*100==>md=Cp*ms/100=6.2*42.2/100=2.6164g KOH

ro2=ms2/Vs2==>ms2=ro2*Vs2=1.075*10=10.75g

Cp=md/ms*100==>md=Cp*ms/100=13.76*10.75/100=1.4792g H3PO4

168g........98g.............212g........54g
3KOH + H3PO4 --> K3PO4 + 3H2O
2.6164g...1.4792g.....x=3.2641g....y=0.83143g

ms=md+mH2O=3.2641+0.83143=4.0955g

Cp=md/ms*100=3.2641/4.0955*100=79.699%

MH3PO4=98g/mol

MKOH=56g/mol

MK3PO4=212g/mol

MH2O=18g/mol