Đáp án:

Giải thích các bước giải:

$1/$

$4Na+O_2\buildrel{{t^o}}\over\longrightarrow$ $2Na_2O$

$2Ca+O_2\buildrel{{t^o}}\over\longrightarrow$ $2CaO$

$2Cu+O_2\buildrel{{t^o}}\over\longrightarrow$ $2CuO$

$3Fe+2O_2\buildrel{{t^o}}\over\longrightarrow$ $Fe_3O_4$

$2Mg+O_2\buildrel{{t^o}}\over\longrightarrow$ $2MgO$

$4Al+3O_2\buildrel{{t^o}}\over\longrightarrow$ $2Al_2O_3$

$2/$

$a,PTPƯ:2KClO_3\buildrel{{t^o}}\over\longrightarrow$ $2KCl+3O_2$

$n_{O_2}=\dfrac{6,72}{22,4}=0,3mol.$

$Theo$ $pt:$ $n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,2mol.$

$⇒m_{KClO_3}=0,2.122,5=24,5g.$

$b,PTPƯ:2KMnO_4\buildrel{{t^o}}\over\longrightarrow$ $K_2MnO_4+MnO_2+O_2$

$n_{O_2}=\dfrac{64}{32}=2mol.$

$Theo$ $pt:$ $n_{KMnO_4}=2n_{O_2}=4mol.$

$⇒m_{KMnO_4}=4.158=632g.$

chúc bạn học tốt!

1/

$4Na+O_2\overset{t^o}\to 2Na_2O$

$2Ca+O_2\overset{t^o}\to 2CaO$

$2Cu+O_2\overset{t^o}\to 2CuO$

$3Fe+2O_2O_2\overset{t^o}\to Fe_3O_4$

$2Mg+O_2\overset{t^o}\to 2MgO$

$4Al+3O_2O_2\overset{t^o}\to 2Al_2O_3$

2/

$a/n_{O_2}=6,72/22,4=0,3mol$

$2KClO_3\overset{t^o}\to 2KCl+3O_2$

Theo pt :

$n_{KClO_3}=2/3.n_{O_2}=2/3.0,3=0,2mol$

$⇒m_{KClO_3}=0,2.122,5=24,5g$

$b/n_{O_2}=64/32=2mol$

$2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$

Theo pt :

$n_{KMnO_4}=2.n_{O_2}=2.2=4mol$

$⇒m_{KMnO_4}=4.158=632g$