\(n_{Ag}=\dfrac{15}{108}=\dfrac{5}{36}\left(mol\right)\)
PTHH: \(C_6H_{12}O_6+Ag_2O\xrightarrow[]{NH_3}2Ag\downarrow+C_6H_{12}O_7\)
\(\dfrac{5}{72}\)<---------------------\(\dfrac{5}{36}\)
\(\rightarrow m_{C_6H_{12}O_6}=\dfrac{5}{72}.180=12,5\left(g\right)\)
\(\rightarrow C\%_{C_6H_{12}O_6}=\dfrac{12,5}{50}.100\%=25\%\)
Không có KQ đúng
