a) $2Mg + O_2 \xrightarrow{t^o} 2MgO$

$n_{Mg} = \dfrac{6,4}{24} = \dfrac{4}{15}(mol)$

Theo PTHH : $n_{MgO} = n_{Mg} = \dfrac{4}{15}(mol)$
$m_{MgO} = \dfrac{4}{15}.40 = 10,67(gam)$

b) $n_{O_2} = \dfrac{1}{2}n_{Mg} = \dfrac{2}{15}(mol)$
$V_{O_2} = \dfrac{2}{15}.22,4 = 2,9867(lít)$

$n_{Ca} = \dfrac{4}{40} = 0,1(mol)$
$n_{O_2} = \dfrac{1,68}{22,4} = 0,075(mol)$
$2Ca + O_2 \xrightarrow{t^o} 2CaO$

Ta thấy :

$n_{Ca} : 2 < n_{O_2} : 1$ nên $O_2$ dư

$n_{CaO} = n_{Ca} = 0,1(mol)$
$CaO + H_2O \to Ca(OH)_2$

$n_{Ca(OH)_2} = n_{CaO} = 0,1(mol)$
$m_{Ca(OH)_2} = 0,1.74 = 7,4(gam)$

a, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(M+H_2SO_4\rightarrow MSO_4+H_2\)

Theo PT: \(n_M=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow M_M=\dfrac{11,2}{0,2}=56\left(g/mol\right)\)

→ M là Fe.

b, Theo PT: \(n_{FeSO_4}=n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\)

⇒ n_{H2SO4 dư} = 0,5.1 - 0,2 = 0,3 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\end{matrix}\right.\)

c, Ta có: \(n_{FeSO_4.7H_2O}=n_{FeSO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeSO_4.7H_2O}=0,2.278=55,6\left(g\right)\)

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

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

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

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

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

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

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

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

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

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