Дано: m(CuCl₂)=80 г;
ω(CuCl₂)=15%;
m(Cu(OH)₂)-?
Решение:
CuCl₂+2KOH=2KCl+Cu(OH)₂↓;
m(CuCl₂)=0,15·80=12 г;
n(CuCl₂)=12⁺135≈0,1 моль;
n(CuCl₂)=n(Cu(OH)₂);
m(Cu(OH)₂)=0,1·98=9,8 г.
Массовая доля = m(соли)/m(h2o)*100%=35гр/750гр*100%=4,67%, если не в процентах 7/150.
1) ch3cooh + naoh = ch3coona + h2o
Ch3coo- + h+ + na+ + oh- = ch3coo- + na+ + h2o
H+ + oh- = h2o
3) 2ch3cooh + cuo = (ch3coo)2cu + h2o
2ch2coo- + 2h+ + cuo = 2ch3coo- + cu+ + h2o
2h+ + cuo = cu2+ + h2o
4) 2ch3cooh + na2co3 = 2ch3coona + co2 + h2o
2ch3coo- + 2h+ + 2na+ + co3 = 2ch3coo- + 2na + co2 + h2o
2h++ co3 = co2 + h2o
5) cuso4 + 2naoh = cu(oh)2 + na2so4
Cu2+ so4+ 2na+ 2oh = cu(oh)2 + 2na + so4
Cu + 2oh = cu(oh)2
6) cu(oh)2 + 2ch3cooh = (ch3coo)2cu + h2o
Cu(oh)2 + 2ch3coo+ 2h = 2ch3coo + cu + h2o
Cu(oh)2 + 2h = cu + h2o
MNaHCO3=w*mраствора/100=2*45/100=0,9 г
mH2O=45-0,9=44,1 г
<span> М.У. NH4Cl + NaOH=NaCl + NH4OH
ПИУ </span>NH4(+) + Cl(-) + Na(+)+OH(-)=Na(+)+Cl(-) + NH4OH
СИУ NH4(+) +OH(-)=NH4OH