\(n_{NaOH}=40.10\%=4\left(g\right)\\ n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaCl}=n_{NaOH}=n_{HCl}=0,1\left(mol\right)\\ C\%_{ddNaCl}=\dfrac{0,1.58,5}{0,1.36,5+40}.100\approx13,402\%\)
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