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a)
\(n_{HCl} = 0,35.2 = 0,7(mol)\\ n_{Mg} = a\ mol ; n_{Al} = b\ mol\)
\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)
Ta có :
\(\left\{{}\begin{matrix}24a+27b=7,5\\2a+3b=0,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
Vậy :
\(\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(gam\right)\\m_{Al}=0,1.27=2,7\left(gam\right)\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}n_{MgCl2}=a=0,2\left(mol\right)\\n_{AlCl3}=b=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{MgCl2}=0,2.95=19\left(gam\right)\\m_{AlCl3}=0,1.133,5=13,35\left(gam\right)\end{matrix}\right.\)
a, Gọi nMg = a; nAl = b (mol)
⇒ 24a + 27b = 7,5 (1)
nHCl = 0.7 (mol)
Mg0 → Mg+2 + 2e
a ..................... 2a
Al0 → Al+3 + 3e
b .................... 3b
2H+ + 2e → H20
0,7 .... 0,7
⇒ 2a + 3b = 0,7 (2)
Từ (1), (2) ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\left(mol\right)\)
⇒ \(\left\{{}\begin{matrix}n_{Mg}=0,2\\n_{Al}=0,1\end{matrix}\right.\left(mol\right)\)
⇒ mMg = 4.8 (g)
và mAl = 2.7 (g)
b, nmagie clorua = nMg = 0,2 (mol)
⇒ mmagie clorua = 19 (g)
nnhôm clorua = nAl = 0,1 (mol)
⇒ mnhôm clorua = 13,35 (g)
mhỗn hợp muối sau phản ứng = 13, 35 + 19 = 32, 25 (g)
Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 27y = 7,8 (1)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
BT e, có: 2x + 3y = 0,8 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{7,8}.100\%\approx30,77\%\\\%m_{Al}\approx69,23\%\end{matrix}\right.\)
b, BTNT Mg và Al, có:
nMgCl2 = nMg = 0,1 (mol)
nAlCl3 = nAl = 0,2 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCl_2}=\dfrac{0,1.95}{0,1.95+0,2.133,5}.100\%\approx26,24\%\\\%m_{AlCl_3}\approx73,76\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)
Ta có: \(m_{tăng}=m_{KL}-m_{H_2}\) \(\Rightarrow m_{H_2}=0,8\left(g\right)\) \(\Rightarrow n_{H_2}=\dfrac{0,8}{2}=0,4\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a_____________________\(\dfrac{3}{2}\)a (mol)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b____________________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}27a+24b=7,8\\\dfrac{3}{2}a+b=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1\cdot24=2,4\left(g\right)\\m_{Al}=5,4\left(g\right)\end{matrix}\right.\)
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https://hoc24.vn/cau-hoi/hoa-tan-hoan-toan-20g-x-gom-mg-va-fe-bang-500ml-dung-dich-hcl-2m-vua-du-thu-duoc-dung-dich-y-cho-y-tac-dung-voi-dung-dich-naoh-du-sau-do-loc-va-thu-duoc-ket-tua-znhiet-phan-hoan-toan-z-trong-khong-kh.2002644222203