a, PT: \(C_6H_5OH+KOH\rightarrow C_6H_5OK+H_2O\)

Ta có: \(m_{KOH}=96.7\%=6,72\left(g\right)\Rightarrow n_{KOH}=\dfrac{6,72}{56}=0,12\left(mol\right)\)

Theo PT: \(n_{C_6H_5OH}=n_{KOH}=0,12\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_6H_5OH}=\dfrac{0,12.94}{16,08}.100\%\approx70,15\%\\\%m_{CH_3OH}\approx29,85\%\end{matrix}\right.\)

b, Trong 8,04 (g) X có: 0,06 mol C₆H₅OH và 0,075 mol CH₃OH

PT: \(2C_6H_5OH+2Na\rightarrow2C_6H_5ONa+H_2\)

\(2CH_3OH+2Na\rightarrow2CH_3ONa+H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_6H_5OH}+\dfrac{1}{2}n_{CH_3OH}=0,0675\left(mol\right)\Rightarrow V_{H_2}=0,0675.22,4=1,512\left(l\right)\)

\(\left\{{}\begin{matrix}n_{C_6H_5ONa}=n_{C_6H_5OH}=0,06\left(mol\right)\\n_{CH_3ONa}=n_{CH_3OH}=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow a=0,06.116+0,075.54=11,01\left(g\right)\)

a, BTNT H, có: \(n_{C_6H_5OH}+n_{C_2H_5OH}=2n_{H_2}=2.\dfrac{2,464}{22,4}=0,22\left(mol\right)\left(1\right)\)

BTNT C, có: \(6n_{C_6H_5OH}+2n_{C_2H_5OH}=n_{CO_2}=n_{CaCO_3}=\dfrac{84}{100}=0,84\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_6H_5OH}=0,1\left(mol\right)\\n_{C_2H_5OH}=0,12\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_6H_5OH}=\dfrac{0,1.94}{0,1.94+0,12.46}.100\%\approx63\%\\\%m_{C_2H_5OH}\approx37\%\end{matrix}\right.\)

b, 3/4 lượng A có n_C6H5OH = 3/4.0,1 = 0,075 (mol)

PT: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH_{\downarrow}+3HBr\)

Theo PT: \(n_{C_6H_2Br_3OH}=n_{C_6H_5OH}=0,075\left(mol\right)\)

\(\Rightarrow m_{\downarrow}=0,075.331=24,825\left(g\right)\)

a, BTNT H, có: \(n_{C_2H_5OH}+n_{C_6H_5OH}=2n_{H_2}=2.\dfrac{1,344}{22,4}=0,12\left(mol\right)\left(1\right)\)

PT: \(C_6H_5OH+KOH\rightarrow C_6H_5OK+H_2O\)

Theo PT: \(n_{C_6H_5OH}=n_{KOH}=0,05.1=0,05\left(mol\right)\left(2\right)\)

Từ (1) và (2) ⇒ n_C2H5OH = 0,12 - 0,05 = 0,07 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,07.46}{0,07.46+0,05.94}.100\%\approx40,66\\\%m_{C_6H_5OH}\approx59,34\%\end{matrix}\right.\)

b, 1/2 lượng A chứa 0,025 (mol) C₆H₅OH

PT: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH+3HBr\)

Theo PT: \(n_{C_6H_2Br_3OH}=n_{C_6H_5OH}=0,025\left(mol\right)\Rightarrow m_{C_6H_2Br_3OH}=0,025.331=8,275\left(g\right)=y\)

\(n_{Br_2}=3n_{C_6H_5OH}=0,075\left(mol\right)\Rightarrow m_{Br_2}=0,075.160=12\left(g\right)\)

\(\Rightarrow C\%_{ddBr_2}=\dfrac{12}{250}.100\%=4,8\%=x\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(2A+2nHCl\rightarrow2ACl_n+nH_2\)

\(\dfrac{0,1}{n}\)                                  0,05       ( mol )

\(M_A=\dfrac{2,3}{\dfrac{0,1}{n}}=23n\)

`@n=1->A` là Natri ( Na )

`@n=2->` Loại

`@n=3->` Loại

Vậy A Natri ( Na )

28) \(M_X=5,625.16=90\left(g/mol\right)\)

Đặt CTPT của X là \(C_xH_yO_z\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{90.53,33\%}{12}=4\\y=\dfrac{90.11,11\%}{1}=10\\z=\dfrac{90-4.12-10}{16}=2\end{matrix}\right.\)

`=> X: C_4H_{10}O_2`

Vì X hòa tan được `Cu(OH)_2` tạo thành dd màu xanh

`=> X` là ancol 2 chức có nhóm `(-OH)` liền nhau
CTCT có thể của X:

\(\left(1\right):CH_2OH-CHOH-CH_2-CH_3\\ \left(2\right):CH_3-CHOH-CHOH-CH_3\)

`=> B`

30)

\(n_{Br_2}=\dfrac{28,8}{160}=0,18\left(mol\right)\)

PTHH: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3OH\downarrow+3HBr\)

              0,06<-----0,18------>0,06

\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_5OH}=9,32-0,06.94=3,68\left(g\right)\\m_{kt}=0,06.331=19,86\left(g\right)\end{matrix}\right.\)

`=> C`

a)

$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$

b)

$n_{Fe} = \dfrac{28}{56} = 0,5(mol)$

$n_{Cl_2} = \dfrac{2,479}{24,79} = 0,1(mol)$

Ta thấy : 

$n_{Fe} :2>n_{Cl_2}:3$ nên $Fe$ dư

$n_{FeCl_3} = \dfrac{2}{3}n_{Cl_2} = \dfrac{0,2}{3}(mol)$
$\Rightarrow m_{FeCl_3} = \dfrac{0,2}{3}.162,5 = 10,83(gam)$

a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

           0,4-->0,5----->0,2

b) \(V_{O_2}=0,5.24,79=12,395\left(l\right)\)

c) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

d) PTHH: \(2Cu+O_2\xrightarrow[]{t^o}2CuO\)

                  1<----0,5

`=> m_{Cu} = 1.64 = 64 (g)`

\(m_{O\left(X\right)}=m.25\%=0,25m\left(g\right)\)

BTNT C: \(n_Z=n_{CO}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\\ M_Z=18.2=36\left(g/mol\right)\)

Áp dụng sơ đồ đường chéo:

\(\dfrac{n_{CO}}{n_{CO_2}}=\dfrac{44-36}{36-28}=\dfrac{1}{1}\Rightarrow n_{CO}=n_{CO_2}=0,03\left(mol\right)\)

Bán phản ứng: \(CO+O\rightarrow CO_2\)

                                 0,03<-0,03

\(\Rightarrow m_Y=m_X-m_{O\left(p\text{ư}\right)}=m-0,03.16=m-0,48\left(g\right)\)

Ta có: \(m_{KL}=m_X-m_O=m-0,25m=0,75m\left(g\right)\)

\(\Rightarrow m_{\left(-NO_3\right)}=m_{mu\text{ối}}-m_{KL}=3,08m-0,75m=2,33m\left(g\right)\\ \Rightarrow n_{\left(-NO_3\right)}=\dfrac{2,33m}{62}\left(mol\right)\)

\(n_{NO}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

BTNT N: \(n_{HNO_3}=n_{\left(-NO_3\right)}+n_{NO}=\dfrac{2,33m}{62}+0,04\left(mol\right)\)

BTNT H: \(n_{H_2O}=\dfrac{1}{2}n_{HNO_3}=\dfrac{2,33m}{124}+0,02\left(mol\right)\)

Theo ĐLBTKL:

\(m_Y+m_{HNO_3}=m_{mu\text{ối}}+m_{NO}+m_{H_2O}\)

\(\Rightarrow m-0,48+63.\left(\dfrac{2,33m}{62}+0,04\right)=3,08m+0,04.30+18.\left(\dfrac{2,33m}{124}+0,02\right)\\ \Leftrightarrow m=9,4777\left(g\right)\)