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![](https://rs.olm.vn/images/avt/0.png?1311)
2KClO3-to\xt->2KCl+3O2
0,1------------------0,1
n KClO3=\(\dfrac{12,25}{122,5}\)=0,1 mol
=>m KCl=0,1.74,5=7,45g
H=\(\dfrac{6,8}{7,45}.100\)=91,275%
b)
2KClO3-to\xt->2KCl+3O2
0,2-------------------------0,3 mol
n O2=\(\dfrac{6,72}{22,4}\)=0,3 mol
H=85%
=>m KClO3=0,2.122,5.\(\dfrac{100}{85}\)=28,82g
c)
2KClO3-to\xt->2KCl+3O2
0,2------------------------0,3
n KClO3=\(\dfrac{24,5}{122,5}\)=0,2 mol
H=80%
=>m O2=0,3.32.\(\dfrac{80}{100}\)=10,4g
![](https://rs.olm.vn/images/avt/0.png?1311)
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo định luật bảo toàn khối lượng ta có:
\(m_{KClO_3\left(pư\right)}=m_{KCl}+m_{O_2}=120+110=230\left(kg\right)\)
Tỉ lệ phần trăm về khối lượng kali clorat phản ứng:
\(\dfrac{230}{260}.100\%\approx88,46\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 )
\(m_{Fe\left(du\right)}=n_{Fe\left(du\right)}.M_{Fe}=\left(0,5-0,225\right).56=15,4g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{Mg} = \dfrac{3,6}{24} = 0,15(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,075(mol)\\ \Rightarrow V_{O_2} = 0,075.22,4 = 1,68(lít)\)
b)
\(2KClO_3 \xrightarrow{t^o,MnO_2} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{2}{3}.0,075 = 0,05(mol)\\ \Rightarrow m_{KClO_3} = 0,05.122,5 = 6,125(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)
b)
\(n_{KClO_3} = \dfrac{36,75}{122,5} = 0,3(mol)\)
Theo PTHH :
\(n_{KCl} = n_{KClO_3} = 0,3(mol)\\ \Rightarrow m_{KCl} = 0,3.74,5 = 22,35(gam)\\ \Rightarrow m_{O_2} = m_{KClO_3} - m_{KCl} = 14,4(gam)\)
c)
Bảo toàn khối lượng :
\(m_{O_2} = 25 - 15,4 = 9,6(gam)\\ \Rightarrow n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{KClO_3} = 0,2.122,5 = 24,5(gam)\\ \%m_{tạp\ chất}= \dfrac{25-24,5}{25}.100\% = 2\%\)
\(a.\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(b.\)
\(n_{KClO_3}=\dfrac{36.75}{122.5}=0.3\left(mol\right)\)
\(\Rightarrow n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}\cdot0.3=0.45\left(mol\right)\)
\(m_{O_2}=0.45\cdot32=14.4\left(g\right)\)
\(m_{KCl}=0.3\cdot74.5=22.35\left(g\right)\)
\(c.\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(a.............a\)
\(m_{Cr}=m_{KCl}+m_{tc}=25-122.5a+74.5a=15.4\left(g\right)\)
\(\Rightarrow a=0.2\)
\(m_{O_2}=\dfrac{3}{2}\cdot0.2\cdot32=9.6\left(g\right)\)
\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)
\(m_{tc}=25-24.5=0.5\left(g\right)\)
\(\%m_{Tc}=\dfrac{0.5}{25}\cdot100\%-2\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nO2 = 16,8 : 22,4 = 0,75 (mol)
2KMnO4 -t--> K2MnO4 + MnO2 + O2
1,5 <-------------------------------0,75(mol)
=> mKClO3 = 1,5 . 122,5 = 183,75 ( g)
C+O2→CO2
1 mol →→ 1 mol
nC=3,6\12=0,3mol→0,3mol
Số mol khí oxi cần có là :
2KClO3to⟶2KCl+3O2
2 mol ---- →→--------- 3 mol
x mol ------←← -------0,3 mol
x=0,3×23=0,2(mol)
Khối lượng KClO3 cần dùng là: 0,2×122,5=24,5(g)