lần sau pạn đăng đúng box nhé
a.\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{9,6}{32}=0,3mol\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
2 2 3 ( mol )
0,2 0,3 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,2.122,5=24,5g\)
b.\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{44,8}{22,4}=2mol\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
2 2 3 ( mol )
4/3 2 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=\dfrac{4}{3}.122,5=\dfrac{490}{3}g\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
a)\(n_{O_2}=\dfrac{9,6}{32}=0,3mol\)
Theo pt: \(\Rightarrow n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\cdot0,3=0,2mol\)
\(m_{KClO_3}=0,2\cdot122,5=24,5g\)
b)\(n_{O_2}=\dfrac{44,8}{22,4}=2mol\)
Theo pt: \(\Rightarrow n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\cdot2=\dfrac{4}{3}mol\)
\(m_{KClO_3}=\dfrac{4}{3}\cdot122,5=163,33g\)
