Trắc nghiệm tổng hợp ôn thi tốt nghiệp THPT môn Hóa Bài 4. Tính chất hoá học của carbohydrate có đáp án

a) Sai

b) Đúng

c) Sai

d) Sai

a) Sai

b) Đúng

c) Đúng

d) Đúng

${({\text{C}}_{6}{\text{H}}_{10}{\text{O}}_{5})}_{\text{n}}\overset{+{\text{nH}}_{2}\text{O},{\text{H}}^{+}}{\rightarrow }{\text{nC}}_{6}{\text{H}}_{12}{\text{O}}_{6};{\text{m}}_{({\text{C}}_{6}{\text{H}}_{10}{\text{O}}_{5}{)}_{n}}=1000\times 0{,}8=800(\text{g})$

${{\rm{n}}_{{{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}}} = {{\rm{n}}_{{{\left( {{{\rm{C}}_6}{{\rm{H}}_{10}}{{\rm{O}}_5}} \right)}_{\rm{n}}}}} = \frac{{800}}{{162}}(\;{\rm{mol}}) \Rightarrow {{\rm{m}}_{{{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}}} = \frac{{800}}{{162}} \times 180 \times \frac{{37{,}5}}{{100}} = 333{,}3(\;{\rm{g}}) \approx 333\;{\rm{g}}$

${({\text{C}}_{6}{\text{H}}_{10}{\text{O}}_{5})}_{\text{n}}\overset{+{\text{nH}}_{2}\text{O},{\text{H}}^{+}}{\rightarrow }{\text{nC}}_{6}{\text{H}}_{12}{\text{O}}_{6}\overset{\text{enzyme}}{\rightarrow }2{\text{nC}}_{2}{\text{H}}_{5}\text{OH}$

${{\rm{m}}_{{{\left( {{{\rm{C}}_6}{{\rm{H}}_{10}}{{\rm{O}}_5}} \right)}_{\rm{n}}}}} = 10 \times \frac{{75}}{{100}} = 7{,}5(\;{\rm{kg}}) \Rightarrow {{\rm{n}}_{{{\left( {{{\rm{C}}_6}{{\rm{H}}_{10}}{{\rm{O}}_5}} \right)}_{\rm{n}}}}} = \frac{{7{,}5}}{{162{\rm{n}}}}({\rm{kmol}})$

${{\rm{n}}_{{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}}} = 2{\rm{n}} \cdot {{\rm{n}}_{{{\left( {{{\rm{C}}_6}{{\rm{H}}_{10}}{{\rm{O}}_5}} \right)}_{\rm{n}}}}} = \frac{{15}}{{162}}({\rm{kmol}}) \Rightarrow {{\rm{m}}_{{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}}} = \frac{{15}}{{162}} \cdot 46 \cdot \frac{{40}}{{100}} = 1{,}704(\;{\rm{kg}})$

$\Rightarrow {{\rm{V}}_{{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}}} = 1{,}704:0{,}789 = 2{,}16(\;{\rm{L}}) \Rightarrow {{\rm{V}}_{{\rm{con }}{{96}^o }}} = 2{,}16:0{,}96 \approx 2{,}25(\;{\rm{L}}).$

${({\text{C}}_{6}{\text{H}}_{10}{\text{O}}_{5})}_{\text{n}}\overset{+{\text{nH}}_{2}\text{O}}{\rightarrow }{\text{nC}}_{6}{\text{H}}_{12}{\text{O}}_{6}\overset{\text{enzyme}}{\rightarrow }2{\text{nC}}_{2}{\text{H}}_{5}\text{OH}$

${{\rm{n}}_{{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}}} = \frac{{100}}{{46}} = 2{,}174({\rm{kmol}}) \Rightarrow {{\rm{n}}_{{{\left( {{{\rm{C}}_6}{{\rm{H}}_{10}}{{\rm{O}}_5}} \right)}_{\rm{n}}}}} = {{\rm{n}}_{{{\rm{C}}_6}{{\rm{H}}_{22}}{{\rm{O}}_6}}} = \frac{1}{{2{\rm{n}}}}{{\rm{n}}_{{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}}} = \frac{{2{,}174}}{{2{\rm{n}}}}({\rm{kmol}})$

${{\rm{m}}_{{{\left( {{{\rm{C}}_6}{{\rm{H}}_{10}}{{\rm{O}}_5}} \right)}_{\rm{n}}}}} = \frac{{2{,}174}}{{2{\rm{n}}}} \times 162{\rm{n}} = 176{,}094(\;{\rm{kg}})$

$\Rightarrow {{\rm{m}}_{{\rm{nguyen lieu }}}} = 176{,}094:0{,}35:0{,}5 \approx 1006(\;{\rm{kg}}).$