Giải thích các bước giải:
Có `n_(C_2H_6)=(5,2)/30=13/75(mol)`
$PT:2C_2H_6+7O_2\to 4CO_2+6H_2O\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{13}{75}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{91}{150}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{26}{75}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,52\text{ }(mol)$
`->m_1=m_(H_2O)=13/25xx18=9,36` `(g)`
`m_(O_2)=91/150xx32=1456/75≈19,413` `(g)`