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\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)
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: \(\left\{{}\begin{matrix}27a+24b=7,8\\\dfrac{3}{2}a+b=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1\cdot24=2,4\left(g\right)\\m_{Al}=5,4\left(g\right)\end{matrix}\right.\)
giúp mình câu này nữa mình tick tiếp cho
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
a) \(\left\{{}\begin{matrix}24.n_{Mg}+27.n_{Al}=6,93\\\dfrac{n_{Mg}}{n_{Al}}=\dfrac{4}{5}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{Mg}=0,12\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)
b)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,12->0,24
2Al + 6HCl --> 2AlCl3 + 3H2
0,15-->0,45
=> nHCl(min) = 0,24 + 0,45 =0,69 (mol)
=> \(V_{dd.HCl\left(min\right)}=\dfrac{0,69}{4}=0,1725\left(l\right)\)
\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.15.....0.3.......................0.15\)
\(m_{Mg}=0.15\cdot24=3.6\left(g\right)\)
\(m_{Cu}=10-3.6=6.4\left(g\right)\)
\(\%Mg=\dfrac{3.6}{10}\cdot100\%36\%\)
\(\%Cu=64\%\)
\(V_{dd_{HCl}}=\dfrac{0.3}{2}=0.15\left(l\right)\)
a) Gọi \(n_{Mg}=4x\left(mol\right)\Rightarrow n_{Al}=5x\left(mol\right)\)
=> \(24.4x+27.5x=6,93\Leftrightarrow x=0,03mol\)
=> \(n_{Mg}=4.0,03=0,12mol\Rightarrow m_{Mg}=2,88g,mAl=6,93-2,88=4,05g\)
b) pt:
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,12 0,24
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,15 0,45
=> nHCl = 0,24+0,45=0,69 mol
=> VHCl = 0,69:4=0,1725 lít
\(Đặt:\left\{{}\begin{matrix}Fe:x\left(mol\right)\\Zn:y\left(mol\right)\end{matrix}\right.\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}56x+65y=5,3\\x+y=0,25\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=1,2\\y=-0,97\end{matrix}\right.\left(vô\:lí\right)\)
Em xem lại đề nha!
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) (1)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (2)
b) Dựa vào đề, ta thấy chắc chắn HCl dư
Ta có: \(\Sigma n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Gọi số mol của Fe là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)
Gọi số mol của Mg là \(b\) \(\Rightarrow n_{H_2\left(2\right)}=b\)
Ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}56a+24b=8\\a+b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=56\cdot0,1=5,6\left(g\right)\\m_{Mg}=24\cdot0,1=2,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{8}\cdot100\%=70\%\\\%m_{Mg}=30\%\end{matrix}\right.\)
c) Theo các PTHH: \(n_{FeCl_2}=n_{MgCl_2}=n_{Fe}=n_{Mg}=0,1mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{muối}=22,2\left(g\right)\)
d) Ta có: \(\Sigma n_{HCl}=\dfrac{500\cdot16\%}{36,5}=\dfrac{160}{73}\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=\dfrac{654}{365}\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=\dfrac{654}{365}\cdot36,5=65,4\left(g\right)\)
Mặt khác: \(m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=507,6\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{12,7}{507,6}\cdot100\%\approx2,5\%\\C\%_{MgCl_2}=\dfrac{9,5}{507,6}\cdot100\%\approx1,87\%\\C\%_{HCl\left(dư\right)}=\dfrac{65,4}{507,6}\cdot100\%\approx12,88\%\end{matrix}\right.\)
\(Đặt:\)
\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
\(m_{hh}=24x+56y=13.6\left(g\right)\\ n_{H_2}=x+y=0.3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0.1\\y=0.2\end{matrix}\right.\)
\(\%Mg=\dfrac{0.1\cdot24}{13.6}\cdot100\%=17.64\%\\ \%Fe=100-17.64=82.36\%\)
\(n_{HCl}=2n_{H_2}=2\cdot0.3=0.6\left(mol\right)\)
\(V_{HCl}=\dfrac{0.6}{2}=0.3\left(l\right)\)
\(m_Y=m_{MgCl_2}+m_{FeCl_2}=0.1\cdot95+0.2\cdot127=34.9\left(g\right)\)
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)