1)
дано
m техн(Na) = 10 g
W(пр)= 5%
V(H2)-?
m чист (Na) = 10 - (10*5% / 100%) = 9.5 g
2Na+2HOH-->2NaOH+H2
M(Na) = 23 g/mol
n(Na) = m/M = 9.5 / 23 = 0.413 mol
n(Na) = n(H2) = 0.413 mol
V(H2) = n*Vm = 0.413*22.4 = 9.25L
ответ 9.25 л
2)
дано
m(ppa KOH) = 450 g
W(KOH) = 25%
CuCL2
m(Cu(OH)2)-?
m(KOH) = 450*25% / 100% = 112.5 g
CuCL2+2KOH-->Cu(OH)2+2KCL
M(KOH) = 56 g/mol
n(KOH) = m/M = 112,5 / 56 = 2 mol
2n(KOH) = n(Cu(OH)2)
n(Cu(OH)2) =2/2 = 1 mol
M(Cu(OH)2) = 98 g/mol
m(Cu(OH)2) = n*M = 1*98 = 98 g
ответ 98 г
Объяснение:
1)
дано
m техн(Na) = 10 g
W(пр)= 5%
V(H2)-?
m чист (Na) = 10 - (10*5% / 100%) = 9.5 g
2Na+2HOH-->2NaOH+H2
M(Na) = 23 g/mol
n(Na) = m/M = 9.5 / 23 = 0.413 mol
n(Na) = n(H2) = 0.413 mol
V(H2) = n*Vm = 0.413*22.4 = 9.25L
ответ 9.25 л
2)
дано
m(ppa KOH) = 450 g
W(KOH) = 25%
CuCL2
m(Cu(OH)2)-?
m(KOH) = 450*25% / 100% = 112.5 g
CuCL2+2KOH-->Cu(OH)2+2KCL
M(KOH) = 56 g/mol
n(KOH) = m/M = 112,5 / 56 = 2 mol
2n(KOH) = n(Cu(OH)2)
n(Cu(OH)2) =2/2 = 1 mol
M(Cu(OH)2) = 98 g/mol
m(Cu(OH)2) = n*M = 1*98 = 98 g
ответ 98 г
Объяснение:
nPbI2=23,05/Mr207+127*2=461)=0,05моль
nPb(NO3)2=nPbI2=0,05
m.Pb(NO3)2= =0,05*Mr207+(14+48)2=331)=0,05*331=16,5г
nNaI=0,05/2=0,025 по уравнению
mNaI=0,025*Mr=0,025+150=3,7
m в-ва=16,5+3,7=20,2
W=m/M=20/300*100%=7%