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Gọi kim loại cần tìm là A
a) PTHH: \(A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\uparrow\)
\(AOH+HCl\rightarrow ACl+H_2O\)
b) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_A=0,2mol\)
\(\Rightarrow M_A=\dfrac{7,8}{0,2}=39\) \(\Rightarrow\) Kim loại cần tìm là Kali
b) Ta có: \(\left\{{}\begin{matrix}n_{KCl}=0,2mol\\n_{HCl\left(pư\right)}=0,2mol\Rightarrow n_{HCl\left(dư\right)}=0,2\cdot20\%=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{KCl}=0,2\cdot74,5=14,9\left(g\right)\\m_{HCl\left(dư\right)}=0,04\cdot36,5=1,46\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{H_2}=2\cdot0,1=0,2\left(g\right)\)
\(\Rightarrow m_{dd}=m_K+m_{ddHCl}-m_{H_2}=7,8+\dfrac{0,24\cdot36,5}{10\%}-0,2=95,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{14,9}{95,2}\cdot100\%\approx15,65\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,46}{95,2}\cdot100\%\approx1,53\%\end{matrix}\right.\)
a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+65b=12,1\) (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Bảo toàn electron: \(2n_{Fe}+2n_{Zn}=2n_{H_2}\) \(\Rightarrow2a+2b=0,4\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{12,1}\cdot100\%\approx46,28\%\\\%m_{Zn}=53,72\%\end{matrix}\right.\)
b)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=n_{Zn}=n_{ZnSO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(p.ứ\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow\Sigma n_{H_2SO_4}=0,2\cdot110\%=0,22\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\\m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,22-0,2\right)\cdot98=1,96\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=211,7\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{15,2}{211,7}\cdot100\%\approx7,18\%\\C\%_{ZnSO_4}=\dfrac{16,1}{211,7}\cdot100\%\approx7,61\%\\C\%_{H_2SO_4}=\dfrac{1,96}{22,4}\cdot100\%\approx0,93\%\end{matrix}\right.\)
Ta có: \(n_{SO_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
BTNT S, có: nH2SO4 = nSO3 = nSO2 = 0,5 (mol)
Mà: mH2SO4 (ban đầu) = 210.10% = 21 (g)
⇒ mH2SO4 (trong X) = 21 + 0,5.98 = 70 (g)
Có: m dd X = 210 + mSO3 = 210 + 0,5.80 = 250 (g)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{70}{250}.100\%=28\%\)
Ta có: \(m_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) \(\Rightarrow m_{H_2}=0,2\cdot2=0,4\left(g\right)\)
\(\Rightarrow m_{dd\left(sau.pư\right)}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}=309,6\left(g\right)\)
\(\Rightarrow a=309,6-300=9,6\left(g\right)\)
a) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
\(n_{HCl}=\dfrac{146.5\%}{36,5}=0,2\left(mol\right)\)
PTHH: CuO + 2HCl --> CuCl2 + H2O
Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\) => CuO hết, HCl dư
=> dd sau phản ứng chứa CuCl2, HCl dư
b)
PTHH: CuO + 2HCl --> CuCl2 + H2O
0,05-->0,1------>0,05
mdd sau pư = 4 + 146 = 150 (g)
\(\left\{{}\begin{matrix}C\%_{CuCl_2}=\dfrac{0,05.135}{150}.100\%=4,5\%\\C\%_{HCldư}=\dfrac{\left(0,2-0,1\right).36,5}{150}.100\%=2,433\%\end{matrix}\right.\)
b)
PTHH: NaOH + HCl --> NaCl + H2O
CuCl2 + 2NaOH --> 2NaCl + Cu(OH)2
0,05--------------------------->0,05
Cu(OH)2 --to--> CuO + H2O
0,05----------->0,05
=> \(a=m_{Cu\left(OH\right)_2}=0,05.98=4,9\left(g\right)\)
=> \(b=m_{CuO}=0,05.80=4\left(g\right)\)
Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Al_2O_3}=y\left(mol\right)\end{matrix}\right.\)
⇒ 27x + 102y = 18,54 (1)
Ta có: \(n_{H_2}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)
Theo ĐLBT mol e, có: 3x = 0,18.2 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,12\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,12.27}{18,54}.100\%\approx17,5\%\\\%m_{Al_2O_3}\approx82,5\%\end{matrix}\right.\)
_ Khi cho hỗn hợp tác dụng với H2SO4 đặc, nóng.
Giả sử: \(n_{SO_2}=a\left(mol\right)\)
Theo ĐLBT mol e, có: 3.0,12 = 2a ⇒ x = 0,18 (mol)
Ta có: \(n_{KOH}=0,36.1=0,36\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{KOH}}{n_{SO_2}}=\dfrac{0,36}{0,18}=2\)
⇒ Pư tạo muối trung hòa K2SO3.
PT: \(SO_2+2KOH\rightarrow K_2SO_3+H_2O\)
___0,18____________0,18 (mol)
\(\Rightarrow C_{M_{K_2SO_3}}=\dfrac{0,18}{0,36}=0,5M\)
Bạn tham khảo nhé!
