Câu 4. Nhiệt phân hoàn toàn 32,67 gam kali clorat KClO3 thì thể tích (đktc) khí oxi thu được là bao nhiêu?
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a,b,
\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)
PTHH: 2KClO3 --to, MnO2--> 2KCl + 3O2
0,1-------------------->0,1------->0,15
\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15.22,4=3,36\left(l\right)\\m_{KCl}=74,5.0,1=7,45\left(g\right)\end{matrix}\right.\)
c, PTHH: 3Fe + 2O2 --to--> Fe3O4
0,225<-0,15------->0,075
=> mFe3O4 = 0,075.232 = 17,4 (g)
a/ Ta có: \(n_{KClO_3}=\dfrac{12.25}{122.5}=0.1\left(mol\right)\)
PTHH:
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)
2 3
0.1 x
\(=>x=\dfrac{0.1\cdot3}{2}=0.15=n_{O_2}\)
\(=>V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)
PTHH: \(KClO_3\underrightarrow{t^o}KCl+\dfrac{3}{2}O_2\)
a) Ta có: \(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
b) PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,15}{2}\) \(\Rightarrow\) Oxi còn dư, Fe p/ứ hết
\(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{15}\left(mol\right)\) \(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}\cdot232\approx15,47\left(g\right)\)
a.\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{12,25}{122,5}=0,1mol\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
2 2 3 ( mol )
0,1 0,15
\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)
b.\(V_{kk}=V_{O_2}.5=3,36.5=16,8l\)
c.\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
3 2 1 ( mol )
0,5 > 0,15 ( mol )
0,225 0,15 ( mol )
\(m_{Fe\left(du\right)}=n_{Fe\left(du\right)}.M_{Fe}=\left(0,5-0,225\right).56=15,4g\)
\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1mol\)
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,1 0,1 0,15 ( mol )
\(m_{KCl}=0,1.74,5=7,45g\)
\(V_{O_2}=0,15.22,4=3,36l\)
a,PTHH: 2Zn+O2−to−>2ZnO2Zn+O2−to−>2ZnO
Bảo toàn khối lượng
⇒mZn=mZnO−mO2=32,4−6,4=26(g)
b,
Ta có: nZn = 6,565=0,1(mol)6,565=0,1(mol)
Theo phương trình, nO2 = 0,12=0,05(mol)0,12=0,05(mol)
=> Thể tích khí Oxi: VO2(đktc) = 0,05 x 22,4 = 1,12 (l)
c,
PTHH:2KClO3to→2KCl+3O2PTHH:2KClO3to→2KCl+3O2
nO2=VO222,4=5,0422,4=0,225(mol)nO2=VO222,4=5,0422,4=0,225(mol)
TheoTheo PTHH,PTHH, tacó:tacó:
nKClO3=23nO2=23.0,225=0,15(mol)nKClO3=23nO2=23.0,225=0,15(mol)
mKClO3=nKClO3.MKClO3=0,15.122,5=18,375(g)mKClO3=nKClO3.MKClO3=0,15.122,5=18,375(g)
Vậy ...
Ko b đúng ko nữa.
2Zn + O2 --> 2ZnO
0,06 <-- 0,03 <----0,06 (mol)
nZnO = \(\dfrac{4,86}{81}\)= 0,06 (mol)
mZn = 0,06 . 65 = 3,9 (g)
VO2 = 0,03 . 22,4 = 0,672 (l)
2KClO3 ----> 2KCl + 3O2
0,02 <------------------- 0,03 (mol)
mKClO3 = 0,02 . (39 + 35,5 + 16.3)
= 2,45 (g)
Kiểm tra lại dùm, thank you
a/ PTHH: 2 KClO3 -to-> 2 KCl + 3 O2
nKCl= 29,8/74,5=0,4(mol)
nKClO3= nKCl=0,4(mol)
=>mKClO3= 0,4.122,5= 49(g)
b) nO2= 3/2. 0,4=0,6(mol)
=> V(O2,đktc)=0,6.22,4=13,44(l)
Ta có: \(n_{KClO_3}=\dfrac{49}{122,5}=0,4\left(mol\right)\)
PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
______0,4______0,4_____0,6 (mol)
\(\Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)
\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
a) 2KClO3 --to--> 2KCl + 3O2
b) \(n_{KClO_3}=\dfrac{36,75}{122,5}=0,3\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,3----------------->0,45
=> V = 0,45.22,4 = 10,08 (l)
nKClO3 = 36,75 : 122,5 = 0,3 (mol)
pthh : 2KClO3 -t--> 2KCl + 3O2
0,3----------------------->0,45 (mol)
=> V= VO2 = 0,45 . 22,4 = 10,08 (L)
nKClO3=0,1(mol)
PTHH: 2 KClO3 -to-> 2 KCl +3 O2
0,1_____________0,1______0,15(mol)
a) mKCl=0,1.74,5=7,45(g)
b) V(O2,đktc)=0,15.22,4=3,36(l)
nKClO3=\(\frac{32,67}{122,5}\)=0,27
PTHH KCLO3➜KCL +O2
theo pt nO2=3/2 nKCLO3=0,405mol
VO2=0,405x22,4=9,072l