NA2SO4 = 23*2+32+16*4=142
C8H16 = 12*8+16*1=112
H2CO3 = 1*2+12+16*3=62
NA2HPO4 = 23*2+1+31+16*4=142
ню(H3PO4)=46/98=0.5моль...............................................................
Дано
M(CxHyOz)=93 g/mol
W(C)=38.7%=0.387
W(O)=51.6%=0.516
W(H)=9.7%=0.097
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CxHyOz-?
CxHyOz/93 = 0.387/12 : 0.097 / 1 : 0.516 / 16
CxHyOz=3:9:3
C3H9O3
M(C3H9O3)=12*3+1*9+16*3=93 g/mol
Дано
m = 120 г
w(HCl) = 35% = 0.35
w(NaOH) = 20% = 0.2
Решение
m(HCl) = mw(HCl) = 120 * 0.35 = 42 г
n(HCl) = m(HCl) / M(HCl) = 42 / (1 + 35.5) = 1.15 моль
NaOH + HCl = NaCl + H2O
n(NaOH) = n(HCl) = 1.15 моль
m(NaOH) = n(NaOH)M(NaOH) = 1.15 * (23 + 16 + 1) = 46 г
m2 = m(NaOH) / w(NaOH) = 230 г
Ответ: 230 г
3)
W=mв-ва/mр-ра *100%
mв-ва=80*5/100=4 г
mр-ра=mв-ва+mH2O=80+50=130 г
W=4/130*100%=3,07%
4)
mв-ва=50*0,1=5 г
m2в-ва=5+8=13 г
mр-ра=50+8=58 г
W=13/58*100%=22,4г