\(BTC:n_{CO_2}=n_{CaCO_3}=\dfrac{30}{100}=0,3mol\)
Lên men rượu:
\(\left(C_6H_{10}O_5\right)_n\rightarrow2nC_2H_5OH+2nCO_2\)
\(\dfrac{0,15}{n}\) 0,3
\(m_{ddgiảm}=m_{ktủa}-m_{CO_2}=12,4\)
\(\Rightarrow m_{CO_2}=30-12,4=17,6g\Rightarrow n_{CO_2}=\dfrac{17,6}{44}=0,4mol\)
Lượng tinh bột phản ứng:
\(m_{tinhbột}=162n\cdot\dfrac{0,15}{n}=24,3g\)
\(H=75\%\Rightarrow m=\dfrac{24,3}{75\%}\cdot100\%=32,4g\)