a) \(n_{CuSO_4}=\dfrac{80.40\%}{160}=0,2\left(mol\right)\)
\(n_{NaOH}=\dfrac{100.20\%}{40}=0,5\left(mol\right)\)
PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => CuSO4 hết, NaOH dư
PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
0,2-------->0,4--------->0,2--------->0,2
=> mNaOH(dư) = (0,5 - 0,4).40 = 4 (g)
b) mdd sau pư = 80 + 100 - 0,2.98 = 160,4 (g)
\(\left\{{}\begin{matrix}C\%_{NaOH\left(dư\right)}=\dfrac{4}{160,4}.100\%=2,5\%\\C\%_{Na_2SO_4}=\dfrac{0,2.142}{160,4}.100\%=17,7\%\end{matrix}\right.\)
a) \(n_{CuSO_4}=\dfrac{80.40\%}{160}=0,2\left(mol\right);n_{NaOH}=\dfrac{100.20\%}{40}=0,5\left(mol\right)\)
PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
ban đầu 0,2 0,5
sau pứ 0 0,1 0,2 0,2
Chất dư là NaOH: \(m_{NaOH\left(dư\right)}=0,1.40=4\left(g\right)\)
b) \(m_{dd.sau.pứ}=80+100-0,2.98=160,4\left(g\right)\)
=> \(C\%_{NaOH}=\dfrac{0,1.40}{160,4}.100\%=2,5\%\\ C\%_{Na_2SO_4}=\dfrac{0,2.142}{160,4}.100\%=17,71\%\)