a) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b) \(n_{CuSO_4}=0,1.1=0,1\left(mol\right)\); \(n_{NaOH}=0,2.2=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) => CuSO4 hết, NaOH dư
PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,1------->0,2--------->0,1----------->0,1
=> \(\left\{{}\begin{matrix}m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\\m_{NaOH\left(dư\right)}=\left(0,4-0,2\right).40=8\left(g\right)\end{matrix}\right.\)
c)
PTHH: \(Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\)
0,1-------->0,1
=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\Rightarrow m_{dd.H_2SO_4}=\dfrac{9,8.100}{20}=49\left(g\right)\)
