Cu(OH)2+ H2SO4 -----> CuSO4+ 2H2O
0.04............0.04.................0.04......0.08
nCu(OH)2=0.04 mol
a) mH2SO4=0.04*98=3.92 g
=>X=\(\dfrac{3.92\cdot100}{200}\)=1.96%
b)mdd=3.92+200=203.92 g
=>C%CuSO4=\(\dfrac{0.04\cdot160\cdot100}{203.92}=3.14\%\)
Cu(OH)2 + H2SO4 → CuSO4 + 2H2O
\(n_{Cu\left(OH\right)_2}=\dfrac{3,92}{98}=0,04\left(mol\right)\)
a) Theo PT: \(n_{H_2SO_4}=n_{Cu\left(OH\right)_2}=0,04\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,04\times98=3,92\left(g\right)\)
\(C\%_{ddH_2SO_4}=\dfrac{3,92}{200}\times100\%=1,96\%\)
Vậy X là 1,96%
b) Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,04\left(mol\right)\)
\(\Rightarrow m_{CuSO_4}=0,04\times160=6,4\left(g\right)\)
\(\Sigma m_{dd}=3,92+200=203,92\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{6,4}{203,92}\times100\%=3,14\%\)
