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nKClO3 = 24.5/122.5 = 0.2 (mol)
2KClO3 -to-> 2KCl + 3O2
0.2_________0.2____0.3
mKCl = 0.2*74.5 = 14.9 (g)
VO2 = 0.3*22.4 = 6.72 (l)
nO2 = 33.6/22.4= 1.5 (mol)
=> nKClO3 = 2/3 * nO2 = 2/3 * 1.5 = 1 (mol)
mKClO3 = 122.5 (g)
\(n_{KClO_3}=\dfrac{24.5}{122.5}=0.2\left(mol\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(0.2.............0.2.........0.3\)
\(m_{KCl}=0.2\cdot74.5=14.9\left(g\right)\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(n_{_{ }O_2}=\dfrac{33.6}{22.4}=1.5\left(mol\right)\)
\(\Rightarrow n_{KClO_3}=1.5\cdot\dfrac{2}{3}=1\left(mol\right)\)
\(m_{KClO_3}=122.5\left(g\right)\)
a) \(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)
b) \(n_{KCl} = n_{KClO_3} = \dfrac{24,5}{122,5} = 0,2(mol)\\ \Rightarrow m_{KCl} = 0,2.74,5 = 14,9(gam)\)
c)
\(n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,3(mol)\\ \Rightarrow V_{O_2} = 0,3.22,4 = 6,72(lít)\)
d)
\(n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{2}{3}.\dfrac{33,6}{22,4} = 1(mol)\\ \Rightarrow m_{KClO_3} = 1.122,5 = 122,5(gam)\)
\(a.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,6\left(mol\right)\\ \Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\\ n_{KCl}=n_{KClO_3}=0,4\left(mol\right)\\ \Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)
\(n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
tỉ lệ 2 : 2 : 3
n(mol) `1/3`<------------`1/3`<-----`0,5`
\(m_{KClO_3}=n\cdot M=\dfrac{1}{3}\cdot\left(39+35,5+16\cdot3\right)\approx40,83\left(g\right)\)
a) PTHH: 2 KClO3 -to-> 2 KCl + 3 O2
nKCl= 14,9/74,5= 0,2(mol)
b) nKClO3=nKCl=0,2(mol)
=>mKClO3=0,2.122,5=24,5(g)
c) nO2=3/2. 0,2=0,3(mol)
=>V(O2,đktc)=0,3.22,4=6,72(l)
\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)
PTHH: 2KClO3 --to--> 2KCl + 3O2 - pư phân huỷ
0,1 0,1 0,15
\(\rightarrow m_{KCl}=0,1.74,5=7,45\left(g\right)\)
a/ PTHH: 2KClO3 =(nhiệt)=> 2KCl + 3O2
nO2 = 9,6 / 32 = 0,3 mol
=> nKClO3 = 0,2 (mol)
=> mKClO3 = 0,2 x 122,5 = 24,5 gam
b/ Cách 1:
nKCl = nKClO3 = 0,2 mol
=> mKCl = 0,2 x 74,5 = 14,9 gam
Cách 2:
Áp dụng định luật bảo toàn khối lượng
=> mKCl = mKClO3 - mO2 = 24,5 - 9,6 = 14,9 gam
\(n_{O_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(0,1..............0,1.......0,15\)
\(m_{KClO_3}=0.1\cdot122.5=12.25\left(g\right)\)
\(m_{KCl}=0,1\cdot74,5=7,45\left(g\right)\)
\(a) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)
b)
Theo PTHH :
\(n_{KCl} = \dfrac{2}{3}n_{O_2} = \dfrac{2}{3}.\dfrac{3,36}{22,4} = 0,1(mol)\\ \Rightarrow m_{KCl} = 0,1.74,5 = 7,45(gam)\)
c)
\(n_{KClO_3} = n_{KCl} = 0,1(mol)\\ \Rightarrow m_{KClO_3} = 0,1.122,5 = 12,25(gam)\)