\(pthh:FeO+H_2\underrightarrow{t^o}Fe+H_2O\) 
                      1       1 
\(m_{Fe}=1.56=56\left(g\right)\)

\(pthh:FeO+H_2\underrightarrow{t^o}Fe+H_2O\) 
                       1       1 
\(m_{Fe}=1.56=56\left(g\right)\)

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ pthh:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ LTL:\dfrac{0,1}{2}>\dfrac{0,2}{6}\) 
=> nhôm oxit  dư 
\(n_{Al_2O_3\left(p\text{ư}\right)}=\dfrac{1}{6}n_{HCl}=\dfrac{1}{30}\left(mol\right)\\ m_{Al\left(d\right)}=\left(0,1-\dfrac{1}{30}\right).102=6,8\left(g\right)\\ n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=\dfrac{1}{15}\left(mol\right)\\ m_{AlCl_3}=\dfrac{1}{15}.133,5=8,9\left(g\right)\)

a. \(n_{Al_2O_3}=\dfrac{10.2}{102}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{7.3}{36,5}=0,2\left(mol\right)\)

PTHH : Al₂O₃ + 6HCl -> 2AlCl₃ + 3H₂O 

              \(\dfrac{0.2}{6}\)        0,2            \(\dfrac{0.4}{6}\)      

Ta thấy : \(\dfrac{0.1}{1}>\dfrac{0.2}{6}\) => Al₂O₃ dư

\(m_{Al_2O_3\left(dư\right)}=\left(0,2-\dfrac{0.2}{6}\right).102=17\left(g\right)\)

b. \(m_{AlCl_3}=\dfrac{0.4}{6}.133,5=8,9\left(g\right)\)

\(n_{CaCl_2.6H_2O}=\dfrac{25}{219}\left(mol\right)\)

=> \(n_{CaCl_2}=\dfrac{25}{219}\left(mol\right)\)

\(C_M=\dfrac{\dfrac{25}{219}}{0,312}=\dfrac{3125}{8541}M\)

\(m_{dd}=1,09.312=340,08\left(g\right)\)

\(C\%=\dfrac{\dfrac{25}{219}.111}{340,08}.100\%=3,726\%\)

\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ pthh:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ LTL:\dfrac{0,2}{1}< \dfrac{0,6}{2}\) 
=> Oxi dư 
\(n_{O_2\left(p\text{ư}\right)}=2n_{CH_4}=0,4\left(mol\right)\\ m_{O_2\left(d\right)}=\left(0,6-0,4\right).32=6,4\left(g\right)\\ n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\\ V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ pthh:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\) '
\(LTL:\dfrac{0,4}{1}>\dfrac{0,6}{2}\) 
=> CH₄ dư 
\(n_{CH_4\left(p\text{ư}\right)}=n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,3\left(mol\right)\\ m_{CH_4\left(d\right)}=\left(0,4-0,3\right).16=1,6\left(g\right)\\ V_{CO_2}=0,3.22,4=6,72\left(l\right)\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\) 
            0,4      1,2         0,4           0,6 
\(V_{H_2}=0,6.22,4=13,44\left(l\right)\\ m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\\ C_{M\left(HCl\right)}=\dfrac{1,2}{0,3}=4M\\ C_{M\left(AlCl_3\right)}=\dfrac{0,4}{0,3}=1,3M\)

mình nghĩ đây là bài toán chất dư (bài toán thừa thiếu) ấy nên số mol của H2 ở đây phải tính theo số mol chất hết chứ nhỉ, đúng k bạn

hỏi 1 lần thôi bn :v

n_Ba=\(\dfrac{27,4}{137}\)=0,2(mol)
pthh:Ba+2H₂O→Ba(OH)₂+H₂↑
     0,2     0,4           0,2               0,2 
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{\text{dd}}=27,4+300-\left(0,2.2\right)=327\left(g\right)\\ m_{Ba\left(OH\right)_2}=171.0,2=34,2\left(g\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{34,2}{327}.100\%=10,5\%\)

\(n_{Ba}=\dfrac{27,4}{137}=0,2\left(mol\right)\\ pthh:Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\uparrow\) 
           0,2     0,4           0,2               0,2 
\(V_{H_2}=0,2.22,4=4,48\left(L\right)\\ m_{\text{dd}}=27,4+300-\left(0,2.2\right)=327\left(g\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{171.0,2}{327}.100\%=10,5\%\)

bạn xem lại hộ mk 27,4g Bari nhé :) 

a)\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)a)(phản ứng phân hủy) 
b) đẩy nước 
c) ngăn chặn không cho chất bị rơi 

hỏi 1 lần thôi bn :)