1)2Fe(OH)3+3H2SO4=Fe2(SO4)3+6H2O
2)2HCl+Na2CO3=2NaCl+H2O+CO2.
3.
Mr(SO)=32+16=48
W(S)=32/48=0.67=67%
W(O)=16/48=0.33=33%
Mr(SO2)=32+2*16=64
W(S)=32/64=0.5=50%
W(O)=32/64=0.5=50%
Mr(SO3)=32+3*16=80
W(S)=32/80=0.4=40%
W(O)=48/80=0.6=60%
H2SO4 ( 2H+ + SO4 2-)
Ca(NO3)2 ((Ca 2+ + 2NO3 - )
Ca(OH)2 (Ca 2+ +2OH - )
HCl H + Cl -
HNO3 H + + NO3(-)
4NH3+5O2 = 4NO+6H2O
n(NH3)= m/M=100/17=5.88 моль
n(NO)=n(NH3)= 5.88 моль
C учетом выхода 85%:
n(NO)= 5.88*0,85= 5 моль
V(NO)=n*Vm=5*22.4=112 л
Ответ: V(NO)=112 л