Hóa học

Tính chất hoá học của `NaHCO_3:`

`-` Tác dụng với dd axit:

`NaHCO_3 + HCl -> NaCl + CO_2 + H_2O`

`-` Tác dụng với dd bazơ (kiềm):

`NaHCO_3 + NaOH -> Na_2CO_3 + H_2O`

`2NaHCO_3 + Ba(OH)_2 -> BaCO_3 + Na_2CO_3 + 2H_2O`

`-` Bị nhiệt phân huỷ:

$2NaHCO_3 \xrightarrow{t^o} Na_2CO_3 + CO_2 + H_2O$

Ta có
`\overline(M)=(16V_1+28V_2)/(V_1+V_2)=26`

`=>10V_1=2V_2`

`=>V_1:V_2=2:10=1:5`

`->C`

`\overline(M)_(hh)=29.1,42=41,18`

Đặt `n_(O_2)=a;n_(CO_2)=b`

Ta có: `(32a+44b)/(a+b)=41,18`

`=>9,18a=2,82b`

`=>a/b=(2,82)/(9,18)=0,31`

Xét các đáp án ta thấy `A` là đáp án đúng nhất `(` Đáp án `D` cũng đúng vì `6:19=12:38)`

\(2CH_4^{\left[A\right]}\xrightarrow[1500^{\circ}C]{lln}C_2H_2^{\left[B\right]}+2H_2^{\left[C\right]}\)

\(C_2H_2^{\left[B\right]}+2AgNO_3+2NH_3\xrightarrow[]{}2NH_4NO_3^{\left[E\right]}+C_2Ag_2^{\left[D\right]}\downarrow\)

\(C_2Ag_2^{\left[D\right]}+2HCl^{\left[F\right]}\xrightarrow[]{}C_2H_2^{\left[B\right]}+2AgCl^{\left[G\right]}\downarrow\)

\(2C_2H_2^{\left[B\right]}\xrightarrow[t^\circ]{NH_4Cl,CuCl}CH\equiv C-CH=CH_2^{\left[H\right]}\)

\(CH\equiv C-CH=CH_2^{\left[H\right]}+H_2^{\left[C\right]}\xrightarrow[t^\circ]{Pd\text{ / }PbCO_3}CH_2=CH-CH=CH_2^{\left[I\right]}\)

\(nCH_2=CH-CH=CH_2\xrightarrow[t^{\circ},p]{xt}\left(-CH_2-CH=CH-CH_2\right)_n^{\left[-\left(I\right)_n-\right]}\)

Gọi tên:

(A): Metan

(B): Axetylen

(C): Hiđro

(D): Bạc axetylua

(E): Amoni nitrat

(F): Axit clohiđric

(G): Bạc clorua

(H): Vinylaxetylen

(I): Buta-1,3-đien

-(I)_n-: Cao su buna

\(A:CH_4;B:C_2H_2;C:H_2\\ D:C_2Ag_2;E:NH_4NO_3\\ F:HCl;G:AgCl\\H:C_4H_4;I:C_4H_6\\ 2CH_4-^{1500^0C,làm.lạnh.nhanh}->C_2H_2+3H_2\\ C_2H_2+2AgNO_3+2NH_3->C_2Ag_2+2NH_4NO_3\\ C_2Ag_2+2HCl->2AgCl+C_2H_2\\ 2C_2H_2-^{CuCl_2,NH_4Cl,100^0C}->H_2C=CH-C\equiv CH\\ H_2C=CH-C\equiv CH+H_2-^{Pd.PbCO_3,t^0}->H_2C=CH-CH=CH_2\\ H_2C=CH-CH=CH_2-^{t^0,p,xt}->-\left(H_2C-CH=CH-CH_2\right)-_n\)

\(2X+2nHCl->2XCl_n+nH_2\\n_{H_2}=\dfrac{4,704}{22,4}=\dfrac{3,78}{X}\cdot\dfrac{n}{2}\\ n:X=\dfrac{1}{9}\\ n=3;X=27\\ X:Al\left(aluminium:nhôm\right) \)

\(N_2+3H_2\leftrightarrow2NH_3\\ n_{NH_3}=a\left(mol\right)\\ n_{sau}=14=9-\dfrac{1}{2}a+6-\dfrac{3}{2}a+a\\ a=1\\ n_{N_2}:1>n_{H_2}:3\\ H=\dfrac{1}{\dfrac{2}{3}.6}.100\%=25\%\)

\(C\%_{AgNO_3\left(60^0C\right)}=\dfrac{m_{AgNO_3\left(bđ\right)}}{2500}=\dfrac{525}{100+525}\\ m_{AgNO_3\left(bđ\right)}=2100\left(g\right)\\ C\%_{AgNO_3\left(10^0C\right)}=\dfrac{170}{270}=\dfrac{2100-m_{AgNO_3tách}}{2500-m_{AgNO_3tách}}\\ m_{AgNO_3tách}=1420\left(g\right)\)