Giúp mk 3 câu này nữa thôi mn ơi . Hứa 5 sao ạ

Đáp án:

`75.D`

`76.A`

`77.D` 

Giải thích các bước giải:

`75)`

`1tan=1000kg`

`n_{(C_6H_{10}O_5)_n}={1000.60\%}/{162n}={100}/{27n}(kmol)`

`(C_6H_{10}O_5)_n+nH_2O`  $\xrightarrow{enzyme}$  `nC_6H_{12}O_6`

`C_6H_{12}O_6`  $\xrightarrow{enzyme}$  `2C_2H_5OH+2CO_2`

`->n_{C_2H_5OH(l t)}={100}/{27n}.2n={200}/{27}(kmol)`

`->m_{C_2H_5OH(t t)}={200}/{27}.46.70\%={6440}/{27}(kg)\approx 238518,5(ml)`

`->V_{C_2H_5OH(t t)}={238518,5}/{0,8}=298148,125(ml)=298,148125(l)`

`->V_{ruou}={29,148125.100}/{70}\approx 425,926(l)`

`76)`

Đặt `n_{glucozo}=x(mol);n_{sacc arozo}=y(mol)`

`->180x+342y=3,51(1)`

`C_{12}H_{22}O_{11}+H_2O`  $\xrightarrow{t^o,H_2SO_4}$  `C_6H_{12}O_6+C_6H_{12}O_6`

`C_6H_{12}O_6`  $\xrightarrow{+AgNO_3/NH_3}$  `2Ag`

`->n_{Ag}=2\sum n_{C_6H_{12}O_6}=2(x+2y)`

`->2x+4y={4,32}/{108}=0,04(2)`

`(1)(2)->x=0,01;y=0,005`

`->m_{sacc arozo}=342.0,005=1,71(g)`

`77)`

Pu đốt cháy của cacbohidrat.

`->n_{CO_2}=n_{O_2}=0,84(mol)`

Bảo toàn C: `6n_{glucozo}+12n_{sacc arozo}=n_{CO_2}`

`->6(n_{glucozo}+2n_{sacc arozo})=0,84`

`->n_{glucozo}+2n_{sacc arozo}=0,14(1)`

`C_{12}H_{22}O_{11}+H_2O`  $\xrightarrow{t^o,H_2SO_4}$  `C_6H_{12}O_6+C_6H_{12}O_6`

`C_6H_{12}O_6`  $\xrightarrow{+AgNO_3/NH_3}$  `2Ag`

`->n_{Ag}=2\sum n_{C_6H_{12}O_6}`

`->n_{Ag}=2(n_{glucozo}+2n_{sacc arozo})`

Thay `(1)` vào:

`->n_{Ag}=0,14.2=0,28(mol)`

`->m=0,28.108=30,24(g)`

75.

`C_6H_{10}O_5->2C_2H_5OH`

`n_{C_6H_{10}O_5}={0, 6.1000}/162=100/27(kmol)`

`H=70%=>n_{C_2H_5OH}=2.{100}/{27}.70%=140/27(kmol)`

`=> V_{C_2H_5OH}={46.140}/{27.0, 8}=8050/27(l)`

`=> V_{dd}=V_r/Đ_r={8050}/{27}.{100}/{70}=425, 926(l)`

`=> D`

76.`n_{Ag}=0, 04(mol)`

Gọi $n_{glucose}=a ; n_{saccazoro}=b$

`=> 180a+342b=3, 51`

`C_{12}H_{22}O_{11}+H_2O->2C_6H_{12}O_6`

`C_6H_{12}O_6->2Ag`

`=> 2a+4b=0, 04`

Giải hệ `=> a=0, 01;b=0, 005`

`=>`$m_{saccarozo}=342.0, 005=1, 71(g)$

`=> A`

77.

$C_6H_{12}O_6+6O_2\xrightarrow{t°}6CO_2+6H_2O$

$C_{12}H_{22}O_{11}+12O_2\xrightarrow{t°}12CO_2+11H_2O$

Gọi $n_{glucose}=a ; n_{saccazoro}=b$

`=> 180a+342b=24, 48`

`6a+12b=0, 84`

`=> a=0, 06; b=0, 04`

Ta có: `n_{Ag}=2a+4b=2.0, 06+4.0, 04=0, 28(mol)`

`=> m_{Ag}=0, 28.108=30, 24g => D`