`2Al + 6HCl -> 2AlCl_3 + 3H_2 \uparrow`
`1/15` `0,2` `0,1` `(mol)`
`n_[H_2]=[3,09875]/[22,4]=0,1(mol)`
`a)m_[Al]=1/15 .27=1,8(g)`
`b)C%_[HCl]=[0,2.36,5]/200 .100=3,65%`
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