a) \(C_nH_{2n-6}+\dfrac{3n-3}{2}O_2\xrightarrow[]{t^o}nCO_2+\left(n-3\right)H_2O\)

b) \(C_xH_yO_z+\dfrac{2x+\dfrac{y}{2}-z}{2}O_2\xrightarrow[]{t^o}xCO_2+\dfrac{y}{2}H_2O\)

a) PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\) 

                \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

Đặt \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{FeCl_3}=2x\left(mol\right)\\n_{MgCl_2}=y\left(mol\right)\end{matrix}\right.\)

Ta lập hệ phương trình: \(\left\{{}\begin{matrix}160x+40y=24\\162,5\cdot2x+95y=52,875\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,3\end{matrix}\right.\)

\(\Rightarrow\%m_{Fe_2O_3}=\dfrac{0,075\cdot160}{24}=50\%=\%m_{MgO}\)

b) Theo PTHH: \(n_{HCl}=6n_{Fe_2O_3}+2n_{MgO}=1,05\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=a=\dfrac{1,05}{0,5}=2,1\left(l\right)\)

PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\uparrow\)

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\) \(\Rightarrow n_{CO_2}=0,1\cdot80\%=0,08\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,08\cdot22,4=1,792\left(l\right)\\m_{CO_2}=0,08\cdot44=3,52\left(g\right)\end{matrix}\right.\)

Bảo toàn khối lượng: \(m_{rắn}=m_{CaCO_3}-m_{CO_2}=6,48\left(g\right)\)

PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)

 Ta có: \(\left\{{}\begin{matrix}n_{K_2SO_3}=\dfrac{39,5}{158}=0,25\left(mol\right)\\n_{HCl}=\dfrac{200\cdot10,95\%}{36,5}=0,6\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,6}{2}\) \(\Rightarrow\) HCl còn dư, tính theo K₂SO₃

 \(\Rightarrow\left\{{}\begin{matrix}n_{SO_2}=0,25\left(mol\right)\\n_{KCl}=0,5\left(mol\right)\\n_{HCl\left(dư\right)}=0,6-0,25\cdot2=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{SO_2}=0,25\cdot22,4=5,6\left(l\right)\\m_{SO_2}=0,25\cdot64=16\left(g\right)\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.pư\right)}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=223,5\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{KCl}=....\\C\%_{HCl\left(dư\right)}=....\end{matrix}\right.\)

Theo thứ tự như đề bài

a) III, II và \(\dfrac{8}{3}\)

b) II, IV, VI

c) II

d) III

ìu :<

kim loại Benzen :)) ?

tan