Câu 4:
Giả sử có 100 (g) hỗn hợp
=> \(m_{H_2}=100.1\%=1\left(g\right)\Rightarrow n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5<-------------------0,5
=> mFe = 0,5.56 = 28 (g)
\(m_{H_2O}=100.21,15\%=21,15\left(g\right)\Rightarrow n_{H_2O}=\dfrac{21,15}{18}=1,175\left(mol\right)\)
Gọi số mol FeO, Fe2O3 là a, b (mol)
=> 72a + 160b = 100 - 28 = 72 (g) (1)
PTHH: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
a--------->a
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b-------------------->3b
=> a + 3b = 1,175 (2)
(1)(2) => a = 0,5 (mol); b = 0,225 (mol)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{28}{100}.100\%=28\%\\\%m_{FeO}=\dfrac{0,5.72}{100}.100\%=36\%\\\%m_{Fe_2O_3}=\dfrac{0,225.160}{100}.100\%=36\%\end{matrix}\right.\)
Câu 5:
(A) là X(HCO3)n
\(m_{X\left(HCO_3\right)_n}=\dfrac{316.6,25}{100}=19,75\left(g\right)\)
PTHH: \(2X\left(HCO_3\right)_n+nH_2SO_4\rightarrow X_2\left(SO_4\right)_n+2nCO_2+2nH_2O\)
Theo PTHH: \(n_{X\left(HCO_3\right)_n}=2.n_{X_2\left(SO_4\right)_n}\)
=> \(\dfrac{19,75}{M_X+61n}=2.\dfrac{16,5}{2.M_X+96n}\)
=> MX = 18n
Xét n = 1 thỏa mãn => MX = 18 (g/mol)
=> X là NH4
(A) là NH4HCO3
\(n_{NH_4HCO_3}=\dfrac{19,75}{79}=0,25\left(mol\right)\)
PTHH: \(NH_4HCO_3+HNO_3\rightarrow NH_4NO_3+CO_2+H_2O\)
0,25-------------------->0,25
(B) có dạng NH4NO3.nH2O (0,25 mol)
\(M_B=\dfrac{47}{0,25}=188\left(g/mol\right)\)
=> n = 6
=> (B) là NH4NO3.6H2O
