a. \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
0,125mol \(\leftarrow\)0,25mol\(\rightarrow\)0,125mol
b. \(n_{NaOH}=\dfrac{10}{40}=0,25mol\\ m_{Cu\left(OH\right)_2}=0,125\cdot98=12,25g\)
c.\(C_{M_{Cu\left(OH\right)_2}}=\dfrac{0,125}{0,1}=1,25M\)
a) CuCl2 + 2NaOH → Cu(OH)2↓ + 2NaCl (1)
0,125<----- 0,25 ----> 0,125 (mol)
b)mCu(OH)2 = 0,125 . 98 = 12,25 g
c) [ CuCl2 ] = 0,125 / 0,1 = 1,25 M