`14.`
`CaCO_3\overset{t^o}{->}CaO+CO_2↑`
`n_{CO_2}=(11,2)/(22,4)=0,5(mol)`
`n_{CaCO_3pư}=n_{CO_2}=0,5(mol)`
`m_{CaCO_3pư}=0,5.100=50(g)`
`H%=50/100 .100=50%`
`=>B`
`15.`
`m(kg)_{CaO}=140.(1-20%)=112(kg)`
`n(kg)_{CaO}=112/56=2(mol)`
`CaO+H_2O->Ca(OH)_2`
`n(kg)_{Ca(OH)_2}=n_{CaO}=2(mol)`
`m(kg)_{Ca(OH)_2}=2.74=148(kg)`
`=>D`
`16.`
`CuO+2HCl->CuCl_2+H_2O`
`Fe_2O_3+6HCl->2FeCl_3+3H_2O`
`n_{HCl}=0,2.3,5=0,7(mol)`
Gọi `a,b` là `n_{CuO},n_{Fe_2O_3}`
`=>m_{hh}=80a+160b=20(1)`
`∑n_{HCl}=2a+6b=0,7(2)`
`(1),(2)=>a=0,05;b=0,1`
`%m_{CuO}=(0,05.80)/20 .100=20%`
`%m_{Fe_2O_3}=(0,1.160)/20 .100=80%`
`=>B`
`\text{___________________________________}`
Đáp án:
`14.B`
`15.D`
`16.B`