pthh: CuSO4 + 2NaOH --> Cu(OH)2 + Na2SO4
mCuSO4=125*19,2%=24g
nCuSO4=24/160=0,15 mol
mNaOH=100*20%=20g
nNaOH=20/40=0,5 mol
có: nNaOH/2 > nCuSO4
-> NaOH dư tính theo CuSO4
-> nCu(OH)2=nCu=0,15 mol
=> mCu(OH)2=m kết tủa=0,15*98=14,7g
nNa2SO4=nCuSO4=0,15 mol
=> mNa2SO4=0,15*142=21,3 g
m dung dịch = m dd cuso4 + mdd naoh - m kết tủa=125+100-14,7=210,3g
=> C% Na2SO4=21,3/210,3 *100%=10,1%
nNaOH dư=0,5-2*0,15=0,2 mol
=> mNaOH dư =0,2*40=8g
=> C% NaOH dư=8/210,3 *100%=3,8%
a) CuSO4 + 2NaOH → Na2SO4 + Cu(OH)2↓
\(m_{CuSO_4}=125\times19,2\%=24\left(g\right)\)
\(\Rightarrow n_{CuSO_4}=\dfrac{24}{160}=0,15\left(mol\right)\)
\(m_{NaOH}=100\times20\%=20\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
Theo PT: \(n_{CuSO_4}=\dfrac{1}{2}n_{NaOH}\)
Theo PT: \(n_{CuSO_4}=\dfrac{3}{10}n_{NaOH}\)
Vì \(\dfrac{3}{10}< \dfrac{1}{2}\) ⇒ NaOH dư
b) Theo pT: \(n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,15\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,15\times98=14,7\left(g\right)\)
c) \(\Sigma m_{dd}saupư=125+100-14,7=210,3\left(g\right)\)
Theo PT: \(n_{NaOH}pư=2n_{CuSO_4}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,5-0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,2\times40=8\left(g\right)\)
\(\Rightarrow C\%_{NaOH}dư=\dfrac{8}{210,3}\times100\%=3,8\%\)
Theo PT: \(n_{Na_2SO_4}=n_{CuSO_4}=0,15\left(mol\right)\)
\(\Rightarrow m_{Na_2SO_4}=0,15\times142=21,3\left(g\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{210,3}\times100\%=10,13\%\)