Bài 8:
a, PTHH: [tex]C_6H_{12}O_6\xrightarrow[menruou]{30-35^oC}2C_2H_5OH+CO_2(1)[/tex]
Theo gt ta có: [tex]n_{C_2H_5OH}=43,2(mol)\Rightarrow V_{C_2H_5OH}=2484(ml)[/tex]
[tex]\Rightarrow D^o=\frac{2484.100}{4600}=54^o[/tex]
b, Ta có: [tex]V_{C_2H_5OH/10,417ml}=5,62518(ml)\\\Rightarrow m_{C_2H_5OH/10,417ml}=45(g)\\\Rightarrow a=5(mg/100mlmau)[/tex]
Bài 7:
a, PTHH: [tex]C_nH_{2n+2-2k}+\frac{3n+1-k}{2}O_2\rightarrow nCO_2+(n-1)H_2O(1)\\C_nH_{2n+2-2k}+2Br_2\rightarrow C_{n}H_{2n+2-k}Br_{2k}(2)[/tex]
Theo (1); (2); gt ta có: [tex]n=k[/tex]
Do đó A là $C_4H_2$
b, PTHH: [tex]C_4H_2+Ag_2O\overset{NH_3}{\rightarrow}Ag_2C_4+H_2O(3)[/tex]
Theo (2) ta có: [tex]n_{C_4H_2}=0,04(mol)[/tex]
Theo (3) ta có: [tex]n_{Ag_2C_4}=0,04(mol)\rightarrow m_{Ag_2C_4}=10,56(g)[/tex]
Bài 9:
PTHH: [tex]C_xH_yCOOC_nH_{2n+1}+NaOH\rightarrow C_xH_yCOONa+C_nH_{2n+1}OH(1)\\C_xH_yCOOC_{n+1}H_{2n+3}+NaOH\rightarrow C_xH_yCOONa+C_{n+1}H_{2n+3}OH(2)[/tex]
Bảo toàn nguyên tô O ta có:
$n_{O/C_xH_yCOOC_{n}C_{2n+1};C_xH_yCOOC_{n+1}H_{2n+3}}=0,3(mol)$
[tex]\Rightarrow n_{C_xH_yCOOC_{n}C_{2n+1};C_xH_yCOOC_nH_{2n+3}}=0,15(mol)[/tex]
[tex]\Rightarrow n_{C_nH_{2n+1}OH;C_{n+1}H_{2n+3}OH}=0,15(mol)\\\Rightarrow \overline{M}_{C_nH_{2n+1}OH;C_{n+1}H_{2n+3}OH}=50,67\\\Rightarrow n=2[/tex]
Mặt khác [tex]n_{C_xH_yCOONa}=0,15(mol)\Rightarrow M_{C_xH_y}=27[/tex]
Do đó 2 este cần tìm là $C_2H_3COOC_2H_5$ và $C_2H_3COOC_3H_7$
Gọi số mol 2 este trong A lần lượt là a;b(mol)
Ta có hệ: [tex]\left\{\begin{matrix} 100a+114b=15,7 & \\ a+b=0,15 & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} a=0,1 & \\ b=0,05 & \end{matrix}\right.\\\Rightarrow \left\{\begin{matrix} m_{C_2H_3COOC_2H_5}=10(g) & \\ m_{C_2H_3COOC_3H_7}=5,7(g) & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \%m_{C_2H_3COOC_2H_5}=63,69\% & \\ \%m_{C_2H_3COOC_3H_7}=36,31\% & \end{matrix}\right.[/tex]
Bài 10:
PTHH: $2Na+2H_2O\rightarrow 2NaOH+H_2(1)\\2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2(2)\\2Na+2HCl\rightarrow 2NaCl+H_2(3)\\2Al+6HCl\rightarrow 2AlCl_3+3H_2(4)\\Fe+2HCl\rightarrow FeCl_2+H_2(5)\\AlCl_3+3NaOH\rightarrow Al(OH)_3+3NaCl(6)\\Al(OH)_3+NaOH\rightarrow NaAlO_2+2H_2O(7)\\FeCl_2+2NaOH\rightarrow Fe(OH)_2+2NaCl(8)\\2Al(OH)_3\rightarrow Al_2O_3+3H_2O(9)\\4Fe(OH)_2+O_2\rightarrow 2Fe_2O_3+4H_2O(10)$
Gọi số mol Na; Al; Fe trong từng phần lần lượt là a;b;c(mol)
Theo (3); (4); (5) và gt ta có:
[tex]n_{HCl/tgpu}=1(mol)\Rightarrow a+3b+2c=1[/tex]
Suy ra $n_{HCl/du}=0,2(mol)$
Theo (6); (8) ta có: [tex]n_{NaOH/(6);(8)}=3b+2c[/tex]
Mà $n_{NaOH}$ tham gia phản ứng muối = 1(mol)
[tex]\Rightarrow n_{NaOH/(7)}=a(mol)[/tex]
+, Xét trường hợp 1: $Al(OH)_3$ bị hoà than hết. Loại vì $n_{Na};n_{Al}$ không thoả mãn phần 1
+, Xét trường hợp 2:
Theo (1); (2) ta có: [tex]\frac{1}{2}a+\frac{3}{2}a=0,2\Rightarrow a=0,1(mol)[/tex]
Mặt khác ta có hệ:
[tex]\left\{\begin{matrix} 3b+2c=0,9 & \\ \frac{b-0,1}{2}.102+\frac{c}{2}.160=17,1 & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} b=0,2 & \\ c=0,15 & \end{matrix}\right.[/tex] [tex]\left\{\begin{matrix} 3b+2c=0,9 & \\ \frac{b-0,1}{2}.102+\frac{c}{2}.160=17,1 & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} b=0,2 & \\ c=0,15 & \end{matrix}\right.[/tex]
Do đó [tex]\left\{\begin{matrix} m=32,2 & \\ m_{Na}=4,6 & \\ m_{Al}=10,8 & \\ m_{Fe}=16,8 & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} \%m_{Na}=14,29\% & \\ \%m_{Al}=33,54\% & \\ \%m_{Fe}=52,17\% & \end{matrix}\right.[/tex]