\(n_{NaOH}=0,25.1=0,25\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,5.0,25=0,125\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+H_2O\\ n_{HCl}=2.n_{Ba\left(OH\right)_2}+n_{NaOH}=2.0,125+0,25=0,5\left(mol\right)\\ V_{ddHCl}=\dfrac{0,5}{2,5}=0,2\left(l\right)=200\left(ml\right)\)
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