Hóa học

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)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{24,5}{2+32+16\cdot4}=0,25\left(mol\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

tỉ lệ:          1    :      1      :       1        :     1

n(mol)     0,2      0,25

n(mol pu)  0,2---->0,2--------->0,2------->0,2

\(\dfrac{n_{Zn}}{1}< \dfrac{n_{H_2SO_4}}{1}\left(\dfrac{0,2}{1}< \dfrac{0,25}{1}\right)\)

=> Zn hết, H_2 SO_4` dư

=> tính theo Zn

\(n_{H_2SO_4}=0,25-0,2=0,05\left(mol\right)\\ m_{H_2SO_4}=n\cdot M=0,05\cdot\left(2+32+16\cdot4\right)=4,9\left(g\right)\\ m_{ZnSO_4}=n\cdot M=0,2\cdot\left(65+32+16\cdot4\right)=32,2\left(g\right)\\ m_{H_2}=n\cdot M=0,2\cdot2=0,4\left(g\right)\)

a) Gọi $n_{CO_2} = a(mol) ; n_{SO_2} = b(mol)$
Ta có : 

$a + b = \dfrac{8,96}{22,4} = 0,4(mol)$

$\dfrac{44a + 64b}{a + b} = 27.2$

Suy ra : a = b = 0,2$
$V_{CO_2} = V_{SO_2} = 0,2.22,4 = 4,48(lít)$

b) Theo PTHH : $n_{K_2SO_3} = n_{SO_2} = 0,2(mol)$

$\Rightarrow m_{K_2SO_3} = 0,2.158 = 31,6(gam)$

Gọi $n_{K_2CO_3} = x(mol) ; n_{Na_2CO_3} = y(mol)$

$\Rightarrow 138x + 106y + 31,6 = 56(1)$

$n_{CO_2} = x + y = 0,2(2)$

Từ (1)(2) suy ra : x = y = 0,1

$m_{K_2CO_3} = 0,1.138 = 13,8(gam) ; m_{Na_2CO_3} = 0,1.106 = 10,6(gam)$

$n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$n_{Na} = \dfrac{4,6}{23} = 0,2(mol)$

$n_{NaOH} = 1.V = V(mol)$

$Na_2O + H_2O \to 2NaOH$
$2Na + 2H_2O \to 2NaOH + H_2$

$n_{NaOH\ tạo thành} = 2n_{Na_2O} + n_{Na} = 0,6(mol)$

$\Rightarrow 0,6 + V = V.5$

$\Rightarrow V = 0,15(lít)$

h%=mtt/mlt*100

ct này bn tính chất tham gia pư

vd Fe+O2→Fe3O4

Fe,O2 la chấttham gia pư

Fe3O4 là sp

a)

$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
b) $n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$n_{KMnO_4} = 2n_{O_2} = 0,6(mol)$
$m_{KMnO_4} = 0,6.158 = 94,8(gam)$

c) $3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$

$n_{Fe_3O_4} = \dfrac{1}{2}n_{O_2} = 0,15(mol)$
$m_{Fe_3O_4} = 0,15.232 = 34,8(gam)$

a)

\(PTHH:2KMnO_4-^{t^o}>K_2MnO_4+MnO_2+O_2\)

b)

tỉ lệ:              2          :          1       :        1       :    1

n(mol)          0,6<---------0,3<------------0,3<-----0,3

\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(m_{KMnO_4}=n\cdot M=0,6\cdot\left(39+55+16\cdot4\right)=94,8\left(g\right)\)

c)

\(PTHH:3Fe+2O_2-^{t^o}>Fe_3O_4\)

tỉ lệ:         3     :     2       :    1

n(mol)    0,45<---0,3------>0,15

\(m_{Fe_3O_4}=n\cdot M=0,15\cdot\left(56\cdot3+16\cdot4\right)=34,8\left(g\right)\)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$

$Zn + 2HCl \to ZnCl_2 + H_2$

$n_{Al} = \dfrac{a}{27} (mol) \Rightarrow n_{H_2} = \dfrac{3}{2}n_{Al} = \dfrac{a}{18}(mol)$

$n_{Zn} = \dfrac{b}{65}(mol) \Rightarrow n_{H_2} = n_{Zn} = \dfrac{b}{65}(mol)$

$\Rightarrow \dfrac{a}{18} = \dfrac{b}{65}$
$\Rightarrow \dfrac{a}{b} = \dfrac{18}{65}$