\(6nC{O_2} + 5n{H_2}O\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{a/s}^{clorophin}} {({C_6}{H_{10}}{O_5})_n} + 6n{O_2}\)
6n.44 162n
?tấn 8,1 tấn
\({m_{C{O_2}}} = \dfrac{{8,1.6n.44}}{{162n}} = 13,2(\ tấn )\)
\({m_{{O_2}}} = \dfrac{{8,1.6n.32}}{{162n}} = 9,6( tấn)\)