a) \(n_{NaOH}=\dfrac{300.8\%}{40}=0,6\left(mol\right)\)
\(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
\(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,3\left(mol\right)\)
=> \(m_{Cu\left(OH\right)_2}=0,3.98=29,4\left(g\right)\)
b) \(Cu\left(OH\right)_2-^{t^o}\rightarrow CuO+H_2O\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,3\left(mol\right)\)
=> \(m_{CuO}=80.0,3=24\left(g\right)\)
c) \(m_{ddsaupu}=300+\dfrac{0,3.160}{16\%}-29,4=570,6\left(g\right)\)
=> \(C\%_{Na_2SO_4}=\dfrac{0,3.142}{570,6}.100=7,47\%\)
mNaOH=24(g) -> nNaOH=0,6(mol)
a) PTHH: 2 NaOH + CuSO4 -> Cu(OH)2 + Na2SO4
0,6____________0,3_______0,3_______0,3(mol)
m(rắn)=mCu(OH)2=0,3.98=29,4(g)
b) Cu(OH)2 -to-> CuO + H2O
0,3__________0,3(mol)
=> m(rắn sau nung)= mCuO=0,3.80=24(g)
PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(n_{NaOH}=\dfrac{300\cdot8\%}{40}=0,6\left(mol\right)\) \(\Rightarrow n_{CuSO_4}=n_{Na_2SO_4}=n_{Cu\left(OH\right)_2}=n_{CuO}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\\m_{Cu\left(OH\right)_2}=0,3\cdot98=29,4\left(g\right)\\m_{CuO}=0,3\cdot80=24\left(g\right)\\m_{ddCuSO_4}=\dfrac{0,3\cdot160}{16\%}=300\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=570,6\left(g\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{42,6}{570,6}\cdot100\%\approx7,47\%\)
