\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ 0,05.........0,1..........0,05..........0,05\left(mol\right)\\ a.C\%_{ddHCl}=\dfrac{0,1.36,5}{200}.100=1,825\%\\ b.m_{Zn}=0,05.65=3,25\left(g\right)\\ c.C\%_{ddZnCl_2}=\dfrac{136.0,05}{3,25+200-0,05.2}.100\approx3,347\%\)

nH2 = 6,72/22,4 = 0,3 (mol)

PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2

nAl = 0,3 : 3 . 2 = 0,2 (mol)

nHCl (Al) = 0,3 . 2 = 0,6 (mol)

mAl = 0,2 . 27 = 5,4 (g)

%mAl = 5,4/25,65 = 20,05%

%mZnO = 100% - 20,05% = 79,95%

mZnO = 25,65 - 5,4 = 20,25 (g)

nZnO = 20,25/81 = 0,25 (mol)

PTHH: ZnO + 2HCl -> ZnCl2 + H2O

nHCl (ZnO) = 0,25 . 2 = 0,5 (mol)

nHCl (đã dùng) = 0,6 + 0,5 = 1,1 (mol)

CMddHCl = 1,1/0,1008 = 10,9M

C% = (10,9 . 36,5)/(10 . 1,19) = 33,43%

Bổ sung Câu 3:

Gọi \(\left\{{}\begin{matrix}n_{Mg}:a\left(mol\right)\\n_{Al}:b\left(mol\right)\end{matrix}\right.\)

\(m_{hh}=24a+27b=5,1\left(1\right)\)

\(V_{HCl}=\frac{600}{1,2}=500\left(ml\right)=0,5\left(l\right)\)

\(\Rightarrow n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

a____2a_______ a ________ a

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b____3b________b ______1,5b

\(n_{HCl}=2a+3b=0,5\left(2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,1.24=2,4\left(g\right)\\m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)

b,

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{2,4}{5,1}.100\%=47,06\%\\\%m_{Al}=100\%-47,06\%=52,94\%\end{matrix}\right.\)

c,

\(\Rightarrow\left\{{}\begin{matrix}CM_{MgCl2}=\frac{0,1}{0,5}=0,2M\\CM_{AlCl3}=\frac{0,1}{0,5}=0,2M\end{matrix}\right.\)

d,

\(m_{dd\left(spu\right)}-m_{kl}+m_{\left(ddHCl\right)}-m_{H2}=5,1+600-\left(0,1+0,1.1,5\right).2=604,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl2}=\frac{0,1.95}{604,6}.100\%=1,57\%\\C\%_{AlCl3}=\frac{0,1.133,5}{604,6}.100\%=2,21\%\end{matrix}\right.\)

Bổ sung câu 4:

a,Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Al}:b\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=\frac{100.25,55\%}{36,5}=0,7\left(mol\right)=2n_{H2}\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

x_______2x______x________x

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

y________3y________y______1,5y

\(\left\{{}\begin{matrix}2a+3b=0,7\left(BTe\right)\\127a+133,5b=38,75\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,2.56.100}{0,2.56+0,1.27}=80,58\%\\\%m_{Al}=100\%-80,58\%=19,42\%\end{matrix}\right.\)

b,\(m_{dd\left(spu\right)}=0,2.56+0,1.27+100-0,7=113,2\left(g\right)\)

a,

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl2}=22,44\%\\\%_{AlCl3}=11,79\%\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{FeCl2}=\frac{0,2}{0,1}=2M\\CM_{AlCl3}=\frac{0,1}{0,1}=1M\end{matrix}\right.\)

2.

2

NaCl+ H2SO4 --> Na2SO4 + 2HCl

0,198 0,099 0,198

nNaCl=12,87/58,5=0,22

nNaCl thực tế =0,22.90%=0,198

VHCl=0,198.22,4=4,4352lit

mNa2SO4=0,099.(23.2+32+16.4)=14,058g

1/ cho 500ml dung dịch AgNO3 1M(D=1,2g/ml) vào 300ml dung dịch HCL 2M(D=1,5g/ml)

a/ tính nồng độ % các chất tạo ra trong dung dịch?

b/ tính nồng độ mol các chất trong dung dịch thu được. Biết cgất rắn chiếm thể tích không đáng kể

\(n_{AgNO_3}=0,5.1=0,5\left(mol\right)\); \(n_{HCl}=0,3.2=0,6\left(mol\right)\)

AgNO3 + HCl ------> AgCl + HNO3

Lập tỉ lệ AgNO3 và HCl: \(\frac{0,5}{1}< \frac{0,6}{1}\)=> Sau phản ứng HCl dư, AgNO3 phản ứng hết

Theo PT: \(n_{AgCl}=n_{HCl\left(p.ứ\right)}=n_{HNO_3}=n_{AgNO_3}=0,5\left(mol\right)\)

m_{dd sau phản ứng}= 1,2.500 + 1,5.300 - 0,5.143,5=978,25(g)

Dung dịch sau phản ứng gồm HCl dư và HNO3

C%_{HCl dư}=\(\frac{\left(0,6-0,5\right).36,5}{978,25}.100=0,373\%\)

C% _HNO3=\(\frac{0,5.63}{978,25}.100=3,22\%\)

\(CM_{HCl\left(dư\right)}=\frac{\left(0,6-0,5\right)}{0,5+0,3}=0,125M\)

\(CM_{HNO_3}=\frac{0,5}{0,5+0,3}=0,625M\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ 0,15.......0,3.........0,15........0,15\left(mol\right)\)

Còn Cu thì không phản ứng với dung dịch HCl.

\(b,\%m_{Zn}=\dfrac{0,15.65}{23}.100\approx42,391\%\\ \Rightarrow\%m_{Cu}\approx57,609\%\\ c,C\%_{ddHCl}=\dfrac{0,3.36,5}{200}.100=5,475\%\\ d,C\%_{ddZnCl_2}=\dfrac{136.0,15}{0,15.65+200-0,15.2}.100\approx9,74\%\)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=0,25\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)

c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{12}\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{12}.342=28,5\left(g\right)\)

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

                   a_____2a______a_____a      (mol)

                \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                    b_____3b_______b_____\(\dfrac{3}{2}\)b         (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27b=36,1\\a+\dfrac{3}{2}b=\dfrac{21,28}{22,4}=0,95\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,5\cdot56=28\left(g\right)\\m_{Al}=8,1\left(g\right)\end{matrix}\right.\)

b+c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl}=2a+3b=1,9\left(mol\right)\\n_{FeCl_2}=0,5\left(mol\right)\\n_{AlCl_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{1,9}{0,2}=9,5\left(M\right)\\C_{M_{FeCl_2}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

\(\text{a) 2Al + 6HCl → 2AlCl3 + 3H2↑}\)

\(\text{2Al + 3H2SO4 → Al2(SO4)3 + 3H2↑}\)

\(\text{Fe + 2HCl → FeCl2 + H2↑}\)

\(\text{Fe + H2SO4 → FeSO4 + H2↑}\)

\(\text{b) nHCl = 2.0,2 = 0,4 mol}\)

\(\text{nH2SO4 = 2,25 . 0,2 = 0,45 mol}\)

Gọi số mol Al, Fe trong hỗn hợp lần lượt là x, y

mhh = 27x + 56y = 19,3(g)

3nAl + 2nFe = nHCl + 2nH2SO4

→ 3x + 2y = 0,4 + 2.0,45 = 1,3

→ x = 0,3; y = 0,2

mAl = 8,1g

mFe = 11,2g

c) nH2 = 1/2nHCl + nH2SO4 = 0,65 (mol)

\(\Rightarrow\text{V H2 = 0,65 . 22,4 = 14,56 (l)}\)

d) nCl = nHCl = 0,4 mol

nSO4 = nH2SO4 = 0,45 mol

\(\text{m muối = mKL + mCl + mSO4}\)

\(\text{= 19,3 + 0,4 . 35,5 + 0,45 . 96}\)

\(\text{ = 76,7(g)}\)

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