13,
`n_(HCl)=0.25·1.5=0.375(mol)`
`PTHH:KOH+HCl->KCl+H_2O`
`Theo pt:n_(KOH)=n_(KCl)=n_(HCl)=0.375(mol)`
`V_(ddKOH)=0.375/2=0.1875(l)`
`V_(ddKCl)=0.2+0.25=0.45(l)`
`C_M(ddKCl)=0.375/0.45=5/6M`
14,
Ta có:
`m_(Al)+m_(Cu)=10`
Vì `Cu` không pư với `H_2SO_4` nên ta có `PTHH`:
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`
`n_(H_2)=6.72/22.4=0.3(mol)`
`Theo pt:n_(Al)=3/2n_(H_2)=0.3·2/3=0.2(mol)`
`m_(Al)=0.2·27=5.4(g)`
`m_(Cu)=10-5.4=4.6(g)`
`Theo pt:n_(H_2SO_4)=n_(H_2)=0.3(mol)`
`m_(ddH_2SO_4)=(0.3·98·100)/20=147(g)`
`\text{Hoangtuan123lienquan}`
