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1)
H2+CuO->Cu+H2O
0,2-----------0,2 mol
nH2=\(\dfrac{4,48}{22,4}\)=0,2 mol
=>m Cu=0,2.64=12,8g
2)
2KClO3-to>2KCl+3O2
0,3----------------------0,45 mol
n KClO3=\(\dfrac{36,75}{122,5}\)=0,3 mol
=>VO2=0,45.22,4=10,08l
3Fe+2O2-to>Fe3O4
0,675--0,45 mol
=>m Fe=0,675.56=37,8g
a, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
b, \(n_{KClO_3}=\dfrac{19,6}{122,5}=0,16\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,24\left(mol\right)\Rightarrow V_{O_2}=0,24.22,4=5,376\left(l\right)\)
c, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{Al}=\dfrac{4}{3}n_{O_2}=0,32\left(mol\right)\Rightarrow m_{Al}=0,32.27=8,64\left(g\right)\)
a) \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
b) số mol của 19,6 g \(KClO_3\) là:
\(n_{KClO_3}=\dfrac{m}{M}=\dfrac{19,6}{122,5}=0,16\left(mol\right)\)
thể tích của khí Oxi (đktc) là:
\(V_{O_2}=n.22,4=0,24.22,4=5,376\left(l\right)\)
c)\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
khối lương Al cần dùng để tác dụng hết Oxi:
\(m_{Al}=n.M=0,32.27=8,64\left(g\right)\)
2KClO3-to>2KCl+3O2
0,2---------------------0,3
4P+5O2-to->2P2O5
--0,3-------0,12 mol
n KClO3=\(\dfrac{24,5}{122,5}=0,2mol\)
=>VO2=0,3.22,4=6,72l
=>m P2O5=0,12.142=17,04g
=>Vkk=6.72.5=33,6l
nKClO3 = 24,5 : 122,5 = 0,2 (mol)
pthh : 2KClO3 -t--> 2KCl +3 O2
0,2---------------------> 0,3(MOL)
VO2 = 0,3 .22,4 = 6,72 (L)
pthh : 4P+5O2-t--> 2P2O 5
0,3---> 0,12 (mol)
=> mP2O5 = 0,12 . 142 = 17,04 (g)
ta co : Vkk = VO2:21% = 6,72 : 21% 32 (l)
$a)2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$b) n_{KClO_3} = \dfrac{73,5}{122,5} = 0,6(mol)$
$n_{KCl} = n_{KClO_3} = 0,6(mol)$
$m_{KCl} = 0,6.74,5 = 44,7(gam)$
$c) n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,9(mol)$
$V_{O_2} = 0,9.22,4 = 20,16(lít)$
\(n_{KClO_3\left(bd\right)}=\dfrac{55,125}{122,5}=0,45\left(mol\right)\)
=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,45.85}{100}=0,3825\left(mol\right)\)
PTHH: 2KClO3 --to,MnO2--> 2KCl + 3O2
0,3825------------------->0,57375
=> \(V_{O_2}=0,57375.22,4=12,852\left(l\right)\)
Câu 3.
a.b.\(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2mol\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,2 0,3 ( mol )
\(V_{O_2}=0,3.22,4=6,72l\)
c.\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,2 < 0,3 ( mol )
0,2 0,1 ( mol )
\(m_{Al_2O_3}=0,1.102=10,2g\)
Câu 4.
a.b.
\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1mol\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,1 0,15 ( mol )
\(V_{O_2}=0,15.22,4=3,36l\)
c.\(n_{Fe}=\dfrac{8,4}{56}=0,15mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,15 < 0,15 ( mol )
0,15 0,05 ( mol )
\(m_{Fe_3O_4}=0,05.232=11,6g\)
\(a.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,6\left(mol\right)\\ \Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\\ n_{KCl}=n_{KClO_3}=0,4\left(mol\right)\\ \Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)
\(n_{KClO_3}=\dfrac{98}{122,5}=0,8\left(mol\right)\)
PTHH : 2KClO3 -> 2KCl + 3O2
0,8 1,2
\(V_{O_2}=1,2.22,4=26,88\left(l\right)\)