\(a,C\%_{CuSO_4\left(bh.t^o_1\right)}=\dfrac{20}{100+20}.100\%=16,67\%\)
Gọi \(n_{CuSO_4.5H_2O}=a\left(mol\right)\left(đk:a>0\right)\)
\(\rightarrow\left\{{}\begin{matrix}n_{CuSO_4}=a\left(mol\right)\\n_{H_2O}=5a\left(mol\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=160a\left(g\right)\\m_{H_2O}=19.5a=90a\left(g\right)\end{matrix}\right.\)
\(m_{H_2O}=\dfrac{134,2}{34,2+100}.100=100\left(g\right)\\ \rightarrow m_{CuSO_4}=134,2-100=34,2\left(g\right)\)
Theo bài ra, ta có:
\(\dfrac{34,2-160a}{100-90a}.100=20\\ \Leftrightarrow a=0,1\left(mol\right)\left(TM\right)\\ \rightarrow m_{CuSO_4.5H_2O}=0,1.250=25\left(g\right)\)
