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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
Câu 1:
a) \(m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)
b) \(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)
c) \(m_{H_2}=0,2.2=0,4\left(g\right)\)
d) \(n_{N_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{N_2}=0,2.28=5,6\left(g\right)\)
e) \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\Rightarrow m_{O_2}=0,3.32=9,6\left(g\right)\)
Câu 2:
a) \(V_{NO_2}=0,25.22,4=5,6\left(mol\right)\)
b) \(V_{CO_2}=0,3.22,4=6,72\left(mol\right)\)
c) \(n_{Cl_2}=\dfrac{3,55}{35,5}=0,1\left(mol\right)\Rightarrow V_{Cl_2}=0,1.22,4=2,24\left(l\right)\)
d) \(n_{N_2O}=\dfrac{1,32}{44}=0,03\left(mol\right)\Rightarrow V_{N_2O}=0,03.22,4=0,672\left(l\right)\)
$2KClO_3 \xrightarrow{t^o} 2KCl +3 O_2$
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)
\(a.\)
\(n_{O_2}=\dfrac{48}{32}=1.5\left(mol\right)\)
\(2KClO_3\underrightarrow{^{t^0}}2KCl+3O_2\)
\(1...............................1.5\)
\(m_{KClO_3}=1\cdot122.5=122.5\left(g\right)\)
\(b.\)
\(n_{O_2}=\dfrac{44.8}{22.4}=2\left(mol\right)\)
\(2KClO_3\underrightarrow{^{t^0}}2KCl+3O_2\)
\(\dfrac{4}{3}.................2\)
\(m_{KClO_3}=\dfrac{4}{3}\cdot122.5=163.3\left(g\right)\)
Bài 2:
a) nK=7,8/39=0,2(mol)
PTHH: 4K + O2 -to-> 2 K2O
nK2O=2/4 . 0,2=0,1(mol) =>mK2O=94.0,1=0,4(g)
nO2=1/4. 0,2=0,05(mol) => V(O2,đktc)=0,05.22,4=1,12(l)
b) PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.0,05=0,1(mol) => mKMnO4=158.0,1=15,8(g)
Bài 3:
nSO2=1,12/22,4=0,05(mol)
nCa(OH)2=5,18/74=0,07(mol)
Vì 1< nCa(OH)2/nSO2=0,07/0,05=1,4<2
=> Sp thu được là muối trung hòa duy nhất, Ca(OH)2 dư
PTHH: Ca(OH)2 + SO2 -> CaSO3 + H2O (1)
0,05_________0,05_____0,05(mol)
b) mCaSO3=0,05.120=6(g)
mCa(OH)2 (dư)=74. (0,07-0,05)= 1,48(g)
\(a)\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3.2 = 0,6(mol)\\ \Rightarrow m_{KMnO_4} = 0,6.158 = 94,8(gam)\\ b)\ n_{KClO_3} = \dfrac{24,5}{122,5} = 0,2(mol)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,3(mol)\\ \Rightarrow m_{O_2} = 0,3.32 = 9,6(gam)\)
a) nO2 = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Pt: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Theo pt: \(n_{KMnO_4}=2nO_2=0,6\left(mol\right)\Rightarrow m_{KMnO_4}=0,6.158=94,8g\)
b) \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,3\left(mol\right)\)
\(Pt:2KClO_3\rightarrow2KCl+3O_2\)
\(nO_2=\dfrac{3}{2}n_{KClO_3}=0,45\left(mol\right)\Rightarrow mO_2=0,45.32=14,4g\)
1)
a) \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
b) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
c) \(n_{H_2O}=\dfrac{15.10^{23}}{6.10^{23}}=2,5\left(mol\right)\)
2)
a) \(n_A=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) => MA = \(\dfrac{3}{0,1}=30\left(g/mol\right)\)
b) \(d_{A/O_2}=\dfrac{30}{32}=0,9375\)
2K + O2 → 2K2O
nK2O = \(\dfrac{18,8}{94}\)= 0,2 mol => nKphản ứng = 0,2 mol , nO2phản ứng = 0,1 mol
VO2 = 0,1.22,4 = 2,24 lít . Oxi chiếm 1/5 thể tích không khí => V không khí = 2,24.5 = 11,2 lít
mK = 0,2.39 = 7,8 gam
2K + O2 → 2K2O
Nếu có 3,36 lít Oxi phản ứng với 0,2 mol kali => nO2 = \(\dfrac{3,36}{22,4}\)= 0,15mol
Ta có tỉ lệ \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\)=> Oxi dư , kali hết .
Khối lượng sp thu được vẫn tính theo kali => nK2O = 0,2 mol
<=> mK2O = 0,2.94 = 18,8 gam