\(n_{KClO_3}=\dfrac{10^6}{122.5}\left(mol\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(\dfrac{10^6}{122.5}.....\dfrac{10^6}{122.5}\)
\(m_{KCl}=\dfrac{10^6}{122.5}\cdot74.5\simeq0.61\left(tấn\right)\)
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