Giúp mk 4 câu này với mn ơi . Hứa 5 sao + câu trả lời hay nhất ạ

$71,C$

$(C_{6}H_{10}O_5)_n \xrightarrow{enzym}nC_6H_{12}O_6 \xrightarrow{enzym}2nC_2H_5OH$

$V_{C_2H_5OH}=\dfrac{46 \cdot 10}{100}=4,6(l)=4600(ml)$

$m_{C_2H_5OH}=4600 \cdot 0,8=3680(gam)$

$n_{C_2H_5OH}=\dfrac{3680}{46}=80(mol)$

$m_{tinh \,\, bột}=\dfrac{80 \cdot 1}{2n} \cdot 162n \cdot \dfrac{1}{80\%}=8100(gam)$

`->`$m_{gạo}=\dfrac{8100}{75\%}=10800(gam)=10,800(kg)$

$72,C$

`-` Cacbonhydrat có công thức tổng quát : $C_n(H_2O)_n$

$n_{O_2}=0,1125(mol)$

$n_{H_2O}=0,1(mol)$

`-` Quy về: $H_2O:0,1;C:x$

`-` Tương quan đốt cháy:

$4x=4n_{O_2}$ hay $x=0,1125(mol)$

`=>`$m=18 \cdot 0,1+12 \cdot 0,1125=3,15(gam)$

$73,A$

$n_{H_2O}=2,4(mol)$

`-` Bảo toàn khối lượng:

$m_{thuốc \,\, súng}=(162+2,4 \cdot 63-18 \cdot 2,4) \cdot 90\%=243(gam)$

$74,B$

$n_{tủa,1}=0,6(mol)$

$Ca(HCO_3)_2+2NaOH \to CaCO_3+Na_2CO_3+2H_2O$

`=>`$n_{tủa,2}=0,1(mol)$

`-` Bảo toàn nguyên tố $C:$

$n_{CO_2}=0,8(mol)$

`=>`$n_{tinh \,\, bột}=\dfrac{0,8}{2n} \cdot \dfrac{1}{75\%}=\dfrac{8}{15n}(mol)$

$m=\dfrac{8}{15n} \cdot 162n=86,4(gam)$

Đáp án:

`71.C`

`72.C`

`73.A`

`74.B` 

Giải thích các bước giải:

`71)`

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

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

`10l=10000ml`

`->V_{C_2H_5OH}={10000.46}/{100}=4600(ml)`

`->n_{C_2H_5OH}={4600.0,8}/{46}=80(mol)`

Do `H=80\%`

`->n_{(C_6H_{10}O_5)_n(l t)}={{80}/{2n}}/{80\%}={50}/n(mol)`

`->m_{(C_6H_{10}O_5)_n(l t)}={50}/{n}.162n=8100(g)`

`->m_{gao}={8100}/{75\%}=10800(g)=10,8(kg)`

`->m=10,8(kg)`

`72)`

Hỗn hợp là cacbohidrat.

`->n_{CO_2}=n_{O_2}={2,52}/{22,4}=0,1125(mol)`

BTKL: `m=1,8+0,1125.44-0,1125.32=3,15(g)`

`73)`

`[C_6H_7O_2(OH)_3]_n+3nHNO_3`  $\xrightarrow{t^o,H_2SO_4\ đ}$  `[C_6H_7O_2(ONO_2)_3]_n+3nH_2O`

Theo PT: `n_{H_2O}=n_{HNO_3}=2,4(mol)`

BTKL: `m_{thuoc}=162+2,4.63-2,4.18=270(g)`

`->m_{thuoc\ t t}=270.90\%=243(g)`

`74)`

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

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

`n_{CaCO_3}={60}/{100}=0,6(mol)`

`n_{Ca(HCO_3)_2}=1/2n_{NaOH}=0,1(mol)`

Bảo toàn C: `n_{CO_2}=0,6+0,1.2=0,8(mol)`

Theo PT: `n_{(C_6H_{10}O_5)_n}={0,8}/{2n}={0,4}/n(mol)`

`->m={{0,4}/{n}.162n}/{75\%}=86,4(g)`