Câu 4:
$n_{C_6H_{12}O_6}=\dfrac{9}{180}=0,05(mol)$
$C_6H_{12}O_6\to 2Ag$
$\to n_{Ag}=2n_{C_6H_{12}O_6}=0,05.2=0,1(mol)$
$\to m_{Ag}=0,1.108=10,8g$
Chọn $A$
Câu 5:
$n_{C_6H_{12}O_6}=\dfrac{16,2}{180}=0,09(mol)$
$\to n_{C_6H_{12}O_6\rm pứ}=0,09.95\%=0,0855(mol)$
$C_6H_{12}O_6\to 2Ag$
$\to n_{Ag}=2n_{C_6H_{12}O_6}=0,171(mol)$
$m_{Ag}=0,171.108=18,468g$
$n_{AgNO_3\text{dùng theo lí thuyết}}=0,09.2=0,18(mol)$
$\to m_{AgNO_3\text{cần lấy}}=0,18.170:95\%=32,21g$
Câu 6:
$n_{Ag}=\dfrac{32,4}{108}=0,3(mol)$
$\to n_{C_6H_{12}O_6}=\dfrac{n_{Ag}}{2}=0,15(mol)$
$\to m_{C_6H_{12}O_6}=0,15.180=27g$
Chọn $C$
