Bài 3
\(a,PTHH:2KClO_3\rightarrow^{t^o}_{MnO_2}2KCl+3O_2\\ b,n_{KClO_3}=\dfrac{4,9}{122,5}=0,04\left(mol\right)\\ \Rightarrow n_{KCl}=0,04\left(mol\right)\\ \Rightarrow m_{KCl\left(p\text{/}ứ\right)}=0,04\cdot74,5=2,98\left(g\right)\\ \Rightarrow H\%=\dfrac{2,235}{2,98}\cdot100\%=75\%\\ c,n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,06\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,06\cdot22,4=1,344\left(l\right)\)
Bài 4:
\(a,n_{Cu\left(NO_3\right)_2}=\dfrac{15,04}{188}=0,08\left(mol\right)\\ PTHH:2Cu\left(NO_3\right)_2\rightarrow^{t^o}2CuO+4NO_2+O_2\uparrow\\ \Rightarrow n_{CuO\left(p\text{/}ứ\right)}=0,08\left(mol\right)\\ \Rightarrow m_{CuO\left(p\text{/}ứ\right)}=0,08\cdot80=6,4\left(g\right)\\ \Rightarrow H\%=\dfrac{6,4}{8,56}\cdot100\%\approx75\%\\ b,n_{NO_2}=0,16\left(mol\right);n_{O_2}=0,04\left(mol\right)\\ \Rightarrow\overline{M_{hh}}=\dfrac{46\cdot0,16+32\cdot0,04}{0,16+0,04}=\dfrac{8,64}{0,2}=43,2\left(g/mol\right)\\ \Rightarrow d_{hh/H_2}=\dfrac{43,2}{2}=21,6\)
