\(KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ m_{giảm}=\dfrac{1}{4}m_A\\ Đặt:x=n_{KMnO_4};y=n_{KClO_3}\left(x,y>0\right)\\ m_{O_2}=m_{giảm}=\dfrac{1}{4}m_A=\dfrac{1}{4}.\left(158x+122,5y\right)\\ Mặt.khác:m_{O_2}=32x+\dfrac{245}{3}y\\ \Rightarrow\dfrac{158x+122,5y}{4}=\dfrac{96x+245y}{3}\\ \Leftrightarrow90x=612,5y\\ \Leftrightarrow\dfrac{x}{y}=\dfrac{612,5}{90}=\dfrac{245}{36}\\ \Rightarrow\%m_{\dfrac{KClO_3}{hhA}}=\dfrac{36.122,5}{36.122,5+245.158}.100\approx10,227\%\)

\(a,PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(2:1:2\left(mol\right)\)

\(0,05:0,025:0,05\left(mol\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

\(b,V_{O_2}=n.22,4=0,025.22,4=0,56\left(l\right)\)

Cho minh hỏi câu c đâu

$n_{Fe(NO_3)_3} = \dfrac{29.41,724\%}{242} = 0,05(mol)$

$m_{dd\ sau\ tách\ tinh\ thể} = 29 - 8,08 = 20,92(gam)$

$n_{Fe(NO_3)_3\ sau\ tách\ tinh\ thể} = \dfrac{20,92.34,704\%}{242} = 0,03(mol)$

$\Rightarrow n_{Fe(NO_3)_3.nH_2O} = 0,05 - 0,03 = 0,02(mol)$

$\Rightarrow M_{Fe(NO_3)_3.nH_2O} = 242 + 18n = \dfrac{8,08}{0,02} = 404$

$\Rightarrow n = 9$

Vậy CT của tinh thể là $Fe(NO_3)_3.9H_2O$ 

Coi $n_{H_2O} = 1(mol)$

Suy ra : $n_H = 2(mol)$

$\Rightarrow n_O = 2.0,875 = 1,75(mol)$

Ta có : $n_O = 4n_{H_2SO_4} + n_{H_2O}$

$\Rightarrow n_{H_2SO_4} = 0,1875(mol)$

$m_{dd} = m_{H_2SO_4} + m_{H_2O} = 0,1875.98 + 1.18 = 36,375(gam)$

$C\%_{H_2SO_4} = \dfrac{0,1875.98}{36,375}.100\% = 50,5\%$

a, PT: \(M_2O_n+2nHCl\rightarrow2MCl_n+nH_2O\)

Ta có: \(n_{HCl}=0,12.2=0,24\left(mol\right)\)

Theo PT: \(n_{M_2O_n}=\dfrac{1}{2n}n_{HCl}=\dfrac{0,12}{n}\left(mol\right)\)

\(\Rightarrow M_{M_2O_n}=\dfrac{4,8}{\dfrac{0,12}{n}}=40n=2M_M+16n\)

\(\Rightarrow M_M=12n\)

Với n = 2 thì M_M = 24 (g/mol)

→ M là Mg.

b, Theo PT: \(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,12\left(mol\right)=n_{MgCl_2.xH_2O}\)

\(\Rightarrow M_{MgCl_2.xH_2O}=\dfrac{24,36}{0,12}=203\left(g/mol\right)\)

⇒ 24 + 71 + 18x = 203 ⇒ x = 6

Vậy: CTHH của tinh thể là MgCl₂.6H₂O.

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,05 (1)

Ta có: \(n_{H_2}=\dfrac{9,52}{22,4}=0,425\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,425\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,2\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)

c, Ta có: m _{dd HCl} = 1,05.500 = 525 (g)

m _{dd sau pư} = m_hh + m _{dd HCl} - m_H2 = 541,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,2.136}{541,2}.100\%\approx5,03\%\\C\%_{AlCl_3}=\dfrac{0,15.133,5}{541,2}.100\%\approx3,7\%\end{matrix}\right.\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,35\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,35.65=22,75\left(g\right)\)

b, Theo PT: \(n_{HCl\left(pư\right)}=2n_{H_2}=0,7\left(mol\right)\)

Mà: axit dùng dư 10% so với lượng pư.

\(\Rightarrow n_{HCl}=0,7+0,7.10\%=0,77\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,77}{2,8}=0,275\left(l\right)\)

c, Ta có: \(D=\dfrac{m}{V}\Rightarrow m_{ddHCl}=0,275.1000.1,04=286\left(g\right)\)

d, Theo PT: \(n_{ZnCl_2}=n_{H_2}=0,35\left(mol\right)\)

Dd X gồm: ZnCl₂ và HCl dư.

n_HCl dư = 07.10% = 0,07 (mol)

Ta có: m _{dd sau pư} = m_Zn + m _{dd HCl} - m_H2 = 308,05 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,35.136}{308,05}.100\%\approx15,452\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,07.36,5}{308,05}.100\%\approx0,829\%\end{matrix}\right.\)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$

Theo PTHH : $n_{CuSO_4} = n_{H_2SO_4} = n_{CuO} = 0,2(mol)$
$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,2.98}{20\%} = 98(gam)$

$\Rightarrow m_{dd\ sau\ pư} = 0,2.80 + 98 = 114(gam)$

Gọi $n_{CuSO_4.5H_2O} =a (mol)$
$\Rightarrow m_{dd\ sau\ tách\ tinh\ thể} = 114 - 250a(gam)$
$n_{CuSO_4\ sau\ tách\ tinh\ thể} = 0,2 - a(mol)$

$\Rightarrow C\%_{CuSO_4} = \dfrac{160(0,2 - a)}{114 - 250a}.100\% = \dfrac{17,4}{17,4 + 100}$

$\Rightarrow a = 0,123(mol)$
$m_{CuSO_4.5H_2O} = 0,123.250 = 30,75(gam)$

GIUPS MK

$n_{Zn} = \dfrac{13}{65} = 0,2(mol)$
$n_{H_2SO_4} = \dfrac{24,5}{98} = 0,25(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$

Ta thấy : 

$n_{Zn} : 1 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư

$n_{ZnSO_4} = n_{H_2} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$

Suy ra : 

$m_{ZnSO_4} = 0,2.161 = 32,2(gam)$
$m_{H_2} = 0,2.2 = 0,4(gam)$
$m_{H_2SO_4\ dư} = (0,25 - 0,2).98 = 4,9(gam)$

PTHH: Zn + H₂SO₄ -> a

Theo ĐLBTKL: 

Theo pt: Zn + H₂SO₄ -> a

=>      \(m_{Zn}+m_{H_2SO_4}=m_a\)

Thay   13   +   24,5     = 37,5 (g)