Gọi $n_{Mg} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 24a + 65b = 15,3(1)$
$Mg + 2HCl \to MgCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : $n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)$

Từ (1)(2) suy ra : $a = \dfrac{81}{410} ; b = \dfrac{21}{205}$

$m_{Mg} = \dfrac{81}{410}.24 = 4,74(gam)$
$m_{Zn} = 15,3 - 4,74 = 10,56(gam)$

a)

$4K + O_2 \xrightarrow{t^o} 2K_2O$
$2Ca + O_2 \xrightarrow{t^o} 2CaO$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$C + O_2 \xrightarrow{t^o} CO_2$

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
b)

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe +3 H_2O$
$PbO + H_2 \xrightarrow{t^o} Pb + H_2O$

c)

$Ca + 2H_2O \to Ca(OH)_2 + H_2$
$2K + 2H_2O \to 2KOH + H_2$
$CaO + H_2O \to Ca(OH)_2$

$M_{CH_4O} = 12 + 4 + 16 = 32$

$M_{C_2H_5ONa} = 12.2 + 5 + 16 + 23 = 68$
$M_{CH_3Cl} = 12 + 3 + 35,5 = 50,5$
$M_{C_2H_4} = 12.2 + 4 = 28$

Chọn đáp án B

Gọi CTTQ hai anken là $C_nH_{2n}$

$C_nH_{2n} + Br_2 \to C_nH_{2n}Br_2$

$m_{anken} = m_{bình\ tăng} = 7,7(gam)$

$n_{anken} = \dfrac{4,48}{22,4} = 0,2(mol)$

$\Rightarrow M_{anken} = 14n = \dfrac{7,7}{0,2} = 38,5$

$\Rightarrow n = 2,75$

Vậy hai anken là $C_2H_4$ và $C_3H_6$

$n_{glucozo} = \dfrac{108}{180} = 0,6(mol)$

$C_6H_{12}O_6 \xrightarrow{t^o} 2CO_2 + 2C_2H_5OH$

$n_{CO_2} = 2n_{glucozo} = 1,2(mol)$
$V_{CO_2} = 1,2.22,4 = 26,88(lít)$

a)

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

$n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$

Theo PTHH : $n_{HCl} = 3n_{Al} = 1,2(mol)$
$\Rightarrow m = \dfrac{1,2.36,5}{14,6\%} = 300(gam)$

b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)$

$M_xO_y + yH_2 \xrightarrow{t^o}xM + yH_2O$

Theo PTHH : $n_{oxit} = \dfrac{1}{y}.n_{H_2} = \dfrac{0,6}{y}(mol)$
$\Rightarrow \dfrac{0,6}{y}(Mx + 16y) = 34,8$

$\Rightarrow \dfrac{x}{y}.M = 42$

Với x = 3 ; y = 4 thì $M = 56(Fe)$

Vậy oxit là $Fe_3O_4$

a) $Fe + H_2SO_4 \to FeSO_4 + H_2$

b) 

Theo PTHH : $n_{H_2SO_4} = n_{H_2} = \dfrac{224}{1000.22,4} = 0,01(mol)$
$C\%_{H_2SO_4} = \dfrac{0,01.98}{100}.100\% = 0,98\%$

$\Rightarrow x = 0,98$

c) $n_{Fe} = n_{H_2} = 0,01(mol)$
$m_{dd\ sau\ pư} = m_{Fe} + m_{dd\ H_2SO_4} - m_{H_2} = 0,01.56 + 100 - 0,01.2 = 100,54(gam)$
$C\%_{FeSO_4} = \dfrac{0,01.152}{100,54}.100\% = 1,51\%$

$n_{Al} = \dfrac{0,54}{27} = 0,02(mol)$

$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
Theo PTHH : 

$n_{O_2} = \dfrac{3}{4}n_{Al} = 0,015(mol)$

$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$n_{KMnO_4\ pư} = 2n_{O_2} = 0,03(mol)$
$n_{KMnO_4\ đã\ dùng} = 0,03 : 80\% = 0,0375(mol)$
$m_{KMnO_4} = 0,0375.158 = 5,925(gam)$

$m_{H_2O} = D.V = 1.120 = 120(gam)$

$m_{dung\ dịch} = m_{đường} + m_{nước} = 10 + 120 = 130(gam)$
$C\%_{đường} = \dfrac{10}{130} = 7,69\%$

a) Hợp chất vô cơ : $Na_2CO_3,CO_2$

b) Hidrocacbon (chỉ chứa C và H) : $CH_4,C_3H_8,C_2H_4$

c) Dẫn xuất hidrocacbon : $CH_3COOH,C_6H_{12}O_6$