M(CuO)=8 n=m/M M=80 n=8/80=0,1
n=N/Na N=n*Na=0,1*6,02*10 23=
6,02*10 23
Дано
m(ppa CHOH) 75 g
W(CHOH) = 40%
m практ(CHOOH) = 40 g
--------------------------
η-?
m(CHOH)= m(ppa CHOH)*W(CHOH) / 100% = 75*40%/100%= 30 g
M(CHOH) = 30 g/mol
n(CHOH)= m/M= 30 / 30 =1 mol
M(CHOOH) = 46 g/mol
n(CHOOH)= m/M= 40/46 = 0. 9 mol
n(CHOH)> n(CHOOH)
CHOH+Cu(OH)2 -->CHOOH+Cu+H2O
n(CHOH)= n(CHOOH) = 1 mol
m(CHOOH)= n*M= 1*46 = 46 g
η= m(практ)/m(теор) * 100% = 40/46*100% = 90%
ответ 90%
1.а
2а
3а
4б
5г
6г
7авд
8бвд
9. 300*0.2=60г
60-15%
х-100%
х=60*100/15=400г
400-300=100г
10.mCO2=4.48/22.4*44=8.8g
mC=8.8*12/44=2.4g
mH=5.4*2/18=0.6g
mO=6.2-2.4-0.6=3.2
vC=2.4/12=0.2mol
vH=0.6/1=0.6mol
vO=3.2/16=0.2mol
0.2/0.6/0.2=1/3/1
CH3O
умножаем все на 2
С2Н5ОН-этанол
М=46г/моль
N=V/Vm
n(H)=5,6л/22,4л/моль=0,25 моль