`#DyHungg`
Ta có: Tỉ lệ khối lượng `CuO : Fe_{2}O_{3}` là `2:3`
`⇒m_{CuO}=32/(2+3)xx2=6,4xx2=12,8(g)`
`⇒m_{Fe_{2}O_{3}}=32-12,8=19,2(g)`
`PTHH:`
`CuO + H_{2} -> Cu + H_{2}O (1)`
`Fe_{2}O_{3} + 3H_{2} -> 2Fe + 3H_{2}O (2)`
`n_{CuO}=m/M=(12,8)/80=0,16 (mol)`
`⇒n_{H_{2} (1)}=n_{CuO}=0,16 (mol)`
`n_{Fe_{2}O_{3}}=(19,2)/160=0,12 (mol)`
`⇒n_{H_{2} (2)}=0,12xx3:1=0,36 (mol)`
`⇒n_{H_{2}}=n_{H_{2} (1)}+n_{H_{2} (2)}=0,16+0,36=0,52(mol)`
`⇒V_{H_{2}}=22,4xx0,52=11,648(l)`
........................
b) `n_{Cu}=n_{CuO}=0,16 (mol)`
`n_{Fe}=0,12xx2:1=0,24 (mol)`
`m_{KL}=m_{Cu}+m_{Fe}=0,16xx64+0,24xx56=23,68(g)`