$$\rm a)$$

Phản ứng xảy ra:

$$\bullet$$ Phần 1:

$$\rm C_6H_{12}O_6+2AgNO_3+NH_3+H_2O\to 2NH_4NO_3+2Ag+C_6H_{12}O_7$$

$$\bullet$$ Phần 2:

$$\rm  (\kern-6pt- C_6H_{10}O_5 -\kern-6pt)_n+nH_2O\xrightarrow{t^°,\ H_2SO_4}nC_6H_{12}O_6\\2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\C_6H_{12}O_6+2AgNO_3+NH_3+H_2O\to 2NH_4NO_3+2Ag+C_6H_{12}O_7$$

$$\rm b)$$

$$\bullet$$ Phần 1:

Sơ đồ phản ứng:

$$\rm C_6H_{12}O_6\xrightarrow{AgNO_3/NH_3}2Ag\\\hspace{0,45cm}0,15\hspace{1cm}\leftarrow\hspace{1,2cm}0,3$$

$$\bullet$$ Phần 2:

Sơ đồ phản ứng:

$$\rm C_6H_{12}O_6\xrightarrow{AgNO_3/NH_3}2Ag\\\hspace{0,45cm}0,96\hspace{1,1cm}\leftarrow\hspace{0,9cm}1,92$$

Ta có: $$\rm n_{C_6H_{12}O_6\ (tinh\ bot)}=0,96-0,15=0,81\ (mol)$$

$$\rm \to n_{C_6H_{12}O_6\ (ban \ dau)}=0,15×2=0,3\ (mol)$$

$$\rm \to n_{tinh \ bot}=2×\dfrac{0,81}{60\%}=2,7\ (mol)$$

$$\rm\to m_X=0,3×180+2,7×162=491,4\ (g)$$

Tham khảo :

$$a)$$

$$C_6H_{12}O_6+2AgNO_3+H_2O+2NH_3 \to 2Ag+2NH_4NO_3+C_6H_{12}O_7$$

$$(C_{6}H_{10}O_5)_n +nH_2O \xrightarrow{H^+,t^o} nC_6H_{12}O_6$$

$$b)$$

@$$P_1:$$ Glucose tan còn tinh bột không tan.

$$C_6H_{12}O_6+2AgNO_3+H_2O+2NH_3 \to 2Ag+2NH_4NO_3+C_6H_{12}O_7$$

=>$$n_{glucose,1p}=\dfrac{n_{Ag}}{2}=0,15(mol)$$

@$$P_2:$$ 

- Gọi $$a(gam)$$ là khối lượng tinh bột của một phần.

$$n_{tinh \,\, bột,1p}=\dfrac{a}{162n}(mol)$$

$$(C_{6}H_{10}O_5)_n +nH_2O \xrightarrow{H^+,t^o} nC_6H_{12}O_6$$

- Hiệu suất là $$60\%:$$

->$$n_{glucose,thủy \,\, phân,1p}= \dfrac{a}{162n} \cdot n \cdot 60\%=\dfrac{a}{270}(mol)$$

$$C_6H_{12}O_6+2AgNO_3+H_2O+2NH_3 \to 2Ag+2NH_4NO_3+C_6H_{12}O_7$$

->$$n_{Ag}=2n_{glucose,1p}+2n_{glucose,thủy \,\, phân,1p}$$

<=>$$1,92=2 \cdot 0,15+2 \cdot \dfrac{a}{270}$$

=>$$a=218,7(gam)$$

->$$m_{1 \,\, phần}=\dfrac{m}{2}=m_{tinh \,\, bột,1p}+m_{glucose,1p}$$

$$=218,7+0,15 \cdot 180=245,7(gam)$$

=>$$m=245,7 \cdot 2 = 491,4(gam)$$