PTHH: `CuO + H_2SO_4 -> CuSO_4 + H_2O`
Theo PT: `n_{H_2SO_4} = n_{CuSO_4} = n_{CuO} = 0,2 (mol)`
`=> m_{ddH_2SO_4} = (0,2.98)/(20\%) = 98 (g)`
`=> m_{dd} = 98 + 0,2.80 = 114 (g)`
`=>` \(\text{dd ban đầu có: }\left\{{}\begin{matrix}m_{CuSO_4}=0,2.160=32\left(g\right)\\m_{H_2O}=114-32=82\left(g\right)\end{matrix}\right.\)
Gọi \(n_{CuSO_4.5H_2O}=a\left(mol\right)\left(ĐK:a< 0\right)\)
`=>` \(\text{tách ra: }\left\{{}\begin{matrix}m_{CuSO_4}=160a\left(g\right)\\m_{H_2O}=5a.18=90a\left(g\right)\end{matrix}\right.\)
`=>` \(\text{dd còn lại: }\left\{{}\begin{matrix}m_{CuSO_4}=32-160a\left(g\right)\\m_{H_2O}=82-90a\left(g\right)\end{matrix}\right.\)
Ta có: \(S_{CuSO_4\left(10^oC\right)}=17,4\left(g\right)\)
`=>` \(\dfrac{32-160a}{82-90a}.100=17,4\)
`<=>` \(a=\dfrac{4433}{36085}\left(mol\right)\left(TM\right)\)
`=>` \(m_{CuSO_4.5H_2O}=\dfrac{4433}{36085}.250\approx30,71\left(g\right)\)
