1) Tinh khoi luong kali pemanganat can dung de dieu che 1,12 lit khi oxi(dktc)
2)Nung nong hoan toan 49g kali clorat, tinh so mililit khi oxi thu duoc(dktc)
1) Tinh khoi luong kali pemanganat can dung de dieu che 1,12 lit khi oxi(dktc)
2)Nung nong hoan toan 49g kali clorat, tinh so mililit khi oxi thu duoc(dktc)
Đáp án:
Giải thích các bước giải:
1)
$2KMnO4–>K2MnO4+MnO2+O2$
$nO2=1,12/22,4=0,05(mol)$
$nKMnO4=2nO2=0,1(mol)$
=>$mKMnO4=0,1.158=15,8(g)$
2)
$2KClO3–>2KCl+3O2$
$nKClO3=49/122,5=0,4(mol)$
$nO2=3/2nKClO3=0,6(mol)$
=>$VO2=0,6.22,4=13,44(l)=13440ml$
$1/n_{O_2}=1,12/22,4=0,05mol$
$2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$
Theo pt :
$n_{KMnO_4}=2.n_{O_2}=2.0,05=0,1mol$
$⇒m_{KMnO_4}=0,1.158=15,8g$
$2/n_{KClO_3}=49/122,5=0,4$
$2KClO_3\overset{t^o}\to 2KCl+3O_2$
Theo pt :
$n_{O_2}=3/2.n_{KClO_3}=3/2.0,4=0,6mol$
$⇒V_{O_2}=0,6.22,4=13,44l=13440ml$