Có: \(\left\{{}\begin{matrix}n_{CO_2}+n_{H_2}=\dfrac{5,6}{22,4}=0,25\\\dfrac{44.n_{CO_2}+2.n_{H_2}}{n_{CO_2}+n_{H_2}}=13,6.2=27,2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{CO_2}=0,15\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=\dfrac{100.25,55}{100.36,5}=0,7\left(mol\right)\)
Sau phản ứng thu được dd D chứa chất tan duy nhất là RCl2
mdd sau pư = 37,5 + 100 - 0,15.44 - 0,1.2 = 130,7 (g)
=> \(m_{RCl_2}=\dfrac{25,44.130,7}{100}=33,25\left(g\right)\)
Bảo toàn Cl: \(n_{RCl_2}=0,35\left(mol\right)\)
=> \(M_{RCl_2}=\dfrac{33,25}{0,35}=95\left(g/mol\right)\)
=> MR = 24 (Mg)
Bảo toàn C: \(n_{MgCO_3}=0,15\left(mol\right)\)
=> \(\%m_{MgCO_3}=\dfrac{0,15.84}{37,5}.100\%=33,6\%\)
Quy đổi X thành \(\left\{{}\begin{matrix}FeO:a\left(mol\right)\\Fe_2O_3:b\left(mol\right)\end{matrix}\right.\)
TN1:
\(n_{Ca\left(OH\right)_2}=0,02.V_1\left(mol\right)\)
PTHH: Ca(OH)2 + CO2 --> CaCO3 + H2O
0,02.V1--->0,02.V1->0,02.V1
CaCO3 + CO2 + H2O --> Ca(HCO3)2
0,03<---0,03<---------------0,03
Ca(HCO3)2 + 2NaOH --> Na2CO3 + CaCO3 + 2H2O
0,03<-----------------------------0,03
=> \(0,02.V_1-0,03=0,05\)
=> V1 = 4 (l)
\(n_{CO_2}=0,02.4+0,03=0,11\left(mol\right)\)
PTHH: FeO + CO --to--> Fe + CO2
a---------------------->a
Fe2O3 + 3CO --to--> 2Fe + 3CO2
b---------------------------->3b
=> a + 3b = 0,11
TN2:
\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{17}{400}=0,0425\left(mol\right)\)
=> a + 2b = 0,0425.2 = 0,085
=> a = 0,035; b = 0,025
m = 0,035.72 + 0,025.160 = 6,52 (g)
PTHH: 2FeO + 4H2SO4 --> Fe2(SO4)3 + SO2 + 4H2O
0,035--------------------------->0,0175
=> V2 = 0,0175.22,4 = 0,392 (l)
TN3:
PTHH: FeO + 2HCl --> FeCl2 + H2O
0,035--------->0,035
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
0,025-------------->0,05
FeCl2 + 2AgNO3 --> Fe(NO3)2 + 2AgCl
0,035-------------------.0,035------>0,07
FeCl3 + 3AgNO3 --> Fe(NO3)3 + 3AgCl
0,05--------------------------------->0,15
Fe(NO3)2 + AgNO3 --> Fe(NO3)3 + Ag
0,035------------------------------>0,035
=> m1 = 143,5.(0,07 + 0,15) + 108.0,035 = 35,35 (g)
