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:
1)2FeCl2 + Cl2 >2FeCl3
2)2Fe+3Cl2===>2FeCl3(500гр.С)
2Fe(OH)3 + 6HCl(конц.)===>2FeCl3+6H2O
Fe2(SO4)3+6NaCl===>2FeCl3+3Na2SO4