\(n_{Ba\left(OH\right)_2}=\dfrac{150.10\%}{171}=\dfrac{5}{57}\left(mol\right)\)
PTHH: `Ba(OH)_2 + 2HCl -> BaCl_2 + 2H_2O`
\(\dfrac{5}{57}\)--------->\(\dfrac{10}{57}\)
\(\Rightarrow m_{HCl}=\dfrac{\dfrac{10}{57}.36,5}{15\%}=\dfrac{7300}{171}\left(g\right)\)
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