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\(n_{KClO_3}=\dfrac{a}{122,5}mol\)
\(n_{KMnO_4}=\dfrac{b}{158}mol\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{a}{122,5}\) \(\dfrac{3a}{245}\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(\dfrac{b}{158}\) \(\dfrac{b}{316}\)
Sau phản ứng các chất còn lại bằng nhau.
\(\Rightarrow m_{KCl}=m_{K_2MnO_4}+m_{MnO_2}\)
Theo hai pt: \(\dfrac{a}{122,5}\cdot74,5=\dfrac{b}{158}\cdot\left(197+87\right)\)
\(\Rightarrow\dfrac{a}{b}=1,48\)
\(\dfrac{V_{O_2\left(KMnO_4\right)}}{V_{O_2\left(KClO_3\right)}}=\dfrac{\dfrac{b}{316}}{\dfrac{3a}{245}}=\dfrac{245b}{948a}=\dfrac{1}{1,48}\cdot\dfrac{245}{948}=0,17\)
a) 2KClO3 \(\underrightarrow{t^o}\) 2KCl + 3O2
mol \(\dfrac{a}{122,5}\rightarrow\dfrac{a}{122,5}\dfrac{3a}{245}\)
2KMnO4 \(\underrightarrow{t^o}\) K2MnO4 + MnO2 + O 2
mol \(\dfrac{b}{158}\rightarrow\dfrac{b}{316}\dfrac{b}{316}\dfrac{b}{316}\)
\(74,5.\dfrac{a}{122,5}=197.\dfrac{b}{316}+87.\dfrac{b}{316}\)
⇔ \(\dfrac{74,5a}{122,5}=\dfrac{71b}{79}\)
⇒ \(\dfrac{a}{b}=\dfrac{71.122,5}{74,5.79}\approx1,478\)
b) \(\dfrac{3a}{245}:\dfrac{b}{316}=\dfrac{3a.316}{245.b}=\dfrac{948}{245}.\dfrac{a}{b}=\dfrac{948}{245}.1,478\approx5,72\)
nKClO3=a/122,5(mol)
nKMnO4=b/158(mol)
\(2KClO3\rightarrow2KCl+3O2\)(1)
a/122,5____a/122,5____1,5a/122,5
\(2KMnO4\rightarrow K2MnO4+MnO2+O2\)(2)
b/158_______b/136_________b/136___b/136
Ta có: 74.a/122,5=197.b/136+87.b/136
Giải ra:
=>a/b~1,478
b)
nO2(1)/nO2(2)=\(\dfrac{1,5a}{\dfrac{122,5}{\dfrac{b}{316}}}\)=\(\dfrac{948}{245}.\dfrac{a}{b}=\dfrac{948}{245}.1,478=5,72\)
2KClO3 ---> 2KCl + 3O2
a/122,5 a/122,5(74,5)
2KMnO4---------------->K2MnO4 +KMnO2 +o2
b/158 b/316(197) b/316(87)
ta có :
a/122,5 *(74,5)=b/316(197)+ b/316(87)
giải hệ pt
b)tương tự đ/s 4,43
Câu 9:
1) nSO2 = 2,24 : 22,4 = 0,1 mol
nO2 = 3,36 : 22,4 = 0,15 mol
mhh = 0,1 . 64 + 0,15 . 32 = 11,2
2. nCO2 = 4,4 : 44 = 0,1 mol
nO2 = 3,2 : 32 = 0,1 mol
Vhh = (0,1 + 0,1 ) . 22,4 = 4,48 l
3. n = \(\frac{3.10^{23}}{6.10^{23}}=0,5mol\)
Câu 10 :
1. C2H5OH + 3O2 -> 2CO2 + 3H2O
2. Tỉ lệ : 1 : 3 : 2 : 3
3.
ta có PTHH: 2KClO3=>2KCl + 3O2
\(\frac{a}{122,5}\)------->\(\frac{a}{122,5}\).74,5-> \(\frac{3a}{2}\).22,4
2KMnO4=>K2MnO4+ MnO2+ O2
\(\frac{b}{158}\)------->\(\frac{b}{2.158}.197\)->\(\frac{b}{2.158}.87\)-> \(\frac{b}{2}.22,4\)
từ 2PT trên ta có : \(\frac{a}{122,5}.74,5=\frac{b}{2.158}.197+\frac{b}{2.158}.87\)
=> a/b=1,78
b) tỉ lệ phản ứng: \(\frac{3a}{2}.22,4:\frac{b}{2}.22,4=\frac{3a}{b}=4,43\)
948/245