Bài 1:
\(n_{H_2SO_4}=0,1.0,25=0,025\left(mol\right)\)
PT: \(H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4+2H_2O\)
Theo PT: \(n_{Ba\left(OH\right)_2}=n_{H_2SO_4}=0,025\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,025}{0,25}=0,1\left(M\right)\)
Bài 2:
Ta có: \(m_{CuCl_2}=137.24,635\%=33,74995\left(g\right)\Rightarrow n_{CuCl_2}=\dfrac{33,74995}{135}\approx0,25\left(mol\right)\)
PT: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_{2\downarrow}\)
Theo PT: \(n_{NaOH}=2n_{CuCl_2}=0,5\left(mol\right)\Rightarrow m_{NaOH}=0,5.40=20\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{20}{10\%}=200\left(g\right)\)
\(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\)
