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\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{6}.122,5=\dfrac{245}{12}\left(g\right)\)
nO2=6,72/22,4=0,3(mol)
PTHH: 2 KClO3 -to->2 KCl +3 O2
Ta có: nKClO3=2/3. 0,3=0,2(mol)
=>mKClO3=0,2.122,5=24,5(g)
a) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,2<-------------------0,3
=> \(m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)
b) \(n_{KClO_3}=\dfrac{490}{122,5}=4\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
4-------------->4---->6
=> \(m_{KCl}=4.74,5=298\left(g\right)\)
=> \(m_{O_2}=6.32=192\left(g\right)\)
2KClO3 \(\underrightarrow{t^o}\) 2KCl + 3O2
a, \(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{KClO_3}=\dfrac{0,3.2}{3}=0,2mol\\ m_{KClO_3}=0,2.122,5=24,5g\)
b, \(n_{KClO_3}=\dfrac{490}{122,5}=4mol\)
\(\Rightarrow m_{KCl}=4.74,5=298g\)
\(n_{O_2}=\dfrac{4.3}{2}=6mol\\ m_{O_2}=6.32=192g\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,02\left(mol\right)\\n_{Fe}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,03\cdot56=1,68\left(g\right)\\V_{O_2}=0,02\cdot22,4=0,448\left(l\right)\end{matrix}\right.\)
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,06 0,04 0,02 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)
\(V_{O_2}=n_{O_2}.22,4=0,04.22,4=0,896l\)
nFe3O4 = 2,32/232 = 0,01 mol
3Fe + 2O2 ➝ Fe3O4
0,03 0,02 0,01 (mol)
a) mFe = 0,03.56 = 1,68 gam
b) VO2 = 0,02.22,4 = 0,448 lít
\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PTHH: \(n_{KClO_3}=\dfrac{0,25.2}{3}\approx0,17\left(mol\right)\)
Vậy muốn điều chế 5,6 lít O2 cần dùng số gam Kali clorat:
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,17.122,5=20,825g\)
\(n_{O2}\)=\(\dfrac{V}{22,4}\)=\(\dfrac{5,6}{22,4}\)=0,25 (mol)
PT : 2KClO3 →to 2KCl + 3O2
số mol: \(\dfrac{1}{6}\) ← \(\dfrac{1}{6}\) ← 0,25
⇒ mKClO3 = n . M = \(\dfrac{1}{6}\) . 122,5 ∼∼ 20,41(g)