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Câu 6.
\(n_{O_2}=\dfrac{16,8}{22,4}=0,75mol\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
1,5 0,75
\(m_{KMnO_4}=1,5\cdot158=237g\)
Câu 7.
\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,04 0,02
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{2}{75}\) 0,04
\(m_{KClO_3}=\dfrac{2}{75}\cdot122,5=\dfrac{49}{15}\approx3,27g\)
\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
a,
\(n_{O2}=\frac{24}{32}=0,75\left(mol\right)\)
\(\rightarrow n_{KMnO_4}=0,75.2=1,5\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=1,5.\left(39+55+16.4\right)=237\left(g\right)\)
b,
\(n_{O2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(\rightarrow n_{KMnO4}=1,5.2=3\left(mol\right)\)
\(\Rightarrow m_{KMnO4}=3.\left(39+55+16,4\right)=474\left(g\right)\)
$2KClO_3 \xrightarrow{t^o} 2KCl +3 O_2$
a) n O2 = 48/32 = 1,5(mol)
n KClO3 = 2/3 n O2 = 1(mol)
m KClO3 = 1.122,5 = 122,5(gam)
b) n O2 = 44,8/22,4 = 2(mol)
n KClO3 = 2/3 n O2 = 4/3 (mol)
m KClO3 = 122,5.4/3 = 163,33(gam)
\(a.\)
\(n_{O_2}=\dfrac{48}{32}=1.5\left(mol\right)\)
\(2KClO_3\underrightarrow{^{t^0}}2KCl+3O_2\)
\(1...............................1.5\)
\(m_{KClO_3}=1\cdot122.5=122.5\left(g\right)\)
\(b.\)
\(n_{O_2}=\dfrac{44.8}{22.4}=2\left(mol\right)\)
\(2KClO_3\underrightarrow{^{t^0}}2KCl+3O_2\)
\(\dfrac{4}{3}.................2\)
\(m_{KClO_3}=\dfrac{4}{3}\cdot122.5=163.3\left(g\right)\)
PTHH: 2KMnO4--->K2MnO4+MnO2+O2
a, nO2=\(\dfrac{33,6}{22,4}=1,5\) mol
Theo pt: nKMnO4=2.nO2=2.1,5= 3 mol
=> mKMnO4= 3.158= 474 (g)
b, nO2= \(\dfrac{24}{32}=0,75\) mol
Theo pt: nKMnO4=2.nO2= 2.0,75= 1,5 mol
=> mKMnO4= 1,5.158= 237 (g)
\(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)
a)
\(n_{O_2} = \dfrac{48}{32} = 1,5(mol)\)
Theo PTHH :
\(n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 1(mol)\\ \Rightarrow m_{KClO_3} = 1.122,5 = 122,5(gam)\)
b)
\(n_{O_2} = \dfrac{44,8}{22,4} = 2(mol) \)
Theo PTHH :
\(n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{4}{3}mol\\ \Rightarrow m_{KClO_3} = \dfrac{4}{3}.122.5 = 163,33(gam)\)
a) PTHH: 2KClO3-->2KCl + 3O2 1mol <---1mol<---1,5mol nO2=48/32=1,5mol=> nKClO3=1 =>mKClO3=1.122,5=122,5 gamvậy cần 122,5 gam KClO3 để điều chế 48 gam khí oxi
b)PTHH: 2KClO3-->2KCl + 3O2 4/3mol <---4/3mol<---2mol nO2=44,8/22,4=2mol=> nKClO3=4/3mol =>mKClO3=4/3.122,5=163,33333gamVậy cần 163,33333gam KClO3 để điều chế 44,8 lít khí oxi ở đktc
a/ Ta co PTHH: \(2KClO_3\) -->2KCl+3\(O_2\)
\(n_{O_2}\)= 48: 32= 1,5 mol
Theo PTHH ta co:
Cu 3 mol \(O_2\) phan ung voi 2 mol \(KClO_3\)
Cu 1,5 mol \(O_2\)phan ung voi 1 mol \(KClO_3\)
\(m_{KClO_3}\)= n.M= 1. 122,5= 122,5 (g)
b/ \(n_{O_2}\)= 44,8: 22,4= 2 (mol)
Theo PTHH ta co:
Cu 3 mol \(O_2\) phan ung voi 2 mol \(KClO_3\)
Cu 2 mol \(O_2\) phan ung voi 1,3 mol \(KClO_3\)
\(m_{KClO_3}\)= n.M= 1,3 . 122.5= 159.25 (g)
\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
a. \(n_{O2}=\frac{24}{32}=0,75\left(mol\right)\)
\(\rightarrow n_{KMnO4}=0,75.2=1,5\)
\(\rightarrow m_{KMnO4}=1,5.\left(39+55+16.4\right)=237\left(g\right)\)
b. \(n_{O2}=\frac{33,6}{22,4}=1,5\)
\(\rightarrow n_{KMnO4}=1,5.2=3\)
\(\rightarrow m_{KMnO4}=3.\left(39+55+16.4\right)=474\left(g\right)\)