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2KClO3-to>2KCl+3O2
0,2---------------------0,3
4P+5O2-to->2P2O5
--0,3-------0,12 mol
n KClO3=\(\dfrac{24,5}{122,5}=0,2mol\)
=>VO2=0,3.22,4=6,72l
=>m P2O5=0,12.142=17,04g
=>Vkk=6.72.5=33,6l
nKClO3 = 24,5 : 122,5 = 0,2 (mol)
pthh : 2KClO3 -t--> 2KCl +3 O2
0,2---------------------> 0,3(MOL)
VO2 = 0,3 .22,4 = 6,72 (L)
pthh : 4P+5O2-t--> 2P2O 5
0,3---> 0,12 (mol)
=> mP2O5 = 0,12 . 142 = 17,04 (g)
ta co : Vkk = VO2:21% = 6,72 : 21% 32 (l)
\(a.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,6\left(mol\right)\\ \Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\\ n_{KCl}=n_{KClO_3}=0,4\left(mol\right)\\ \Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)
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)
a) 2KClO3 (7/75 mol) \(\underrightarrow{t^o}\) 2KCl (7/75 mol) + 3O2\(\uparrow\) (0,14 mol).
b) Số mol khí oxi là 4,48/32=0,14 (mol).
Khối lượng kali clorat cần dùng là 7/75.122,5=343/30 (g).
Khối lượng chất rắn thu được là 7/75.74,5=1043/150 (g).
\(a,PTHH:2KClO_3\underrightarrow{t^o,MnO_2}2KCl+3O_2\uparrow\\ b,n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ Theo.pt:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ m_{KClO_3}=\dfrac{2}{15}.122,5=\dfrac{49}{3}\left(g\right)\)
a.\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,6 0,45 0,3 ( mol )
\(m_{Al}=0,6.27=16,2g\)
\(V_{O_2}=0,45.22,4=10,08l\)
\(V_{kk}=10,08.5=50,4l\)
b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,3 0,45 ( mol )
\(m_{KClO_3}=0,3.122,5=36,75g\)
c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,3 0,45 ( mol )
\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)
\(m_{KClO_3}=0,4.122,5=49g\)
PTHH: \(Cu_2S+2O_2\xrightarrow[]{t^o}2CuO+SO_2\)
a) Ta có: \(n_{Cu_2S}=\dfrac{100}{160}=0,625\left(mol\right)\) \(\Rightarrow n_{O_2\left(lýthuyết\right)}=1,25\left(mol\right)\)
\(\Rightarrow V_{O_2\left(thực\right)}=\dfrac{1,25\cdot22,4}{96\%}\approx29,17\left(l\right)\)
b) Sửa đề: "Tính khối lượng KMnO4 để hấp thụ hết SO2"
PTHH: \(5SO_2+2KMnO_4+2H_2O\rightarrow K_2SO_4+2MnSO_4+2H_2SO_4\)
Ta có: \(n_{SO_2\left(thực\right)}=n_{Cu_2S}\cdot96\%=0,6\left(mol\right)\)
\(\Rightarrow n_{KMnO_4}=0,24\left(mol\right)\) \(\Rightarrow m_{KMnO_4}=0,24\cdot158=37,92\left(g\right)\)
c) PTHH: \(SO_2+\dfrac{1}{2}O_2\xrightarrow[V_2O_5]{t^o}SO_3\)
Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{SO_2}=0,3\left(mol\right)\) \(\Rightarrow V_{kk}=\dfrac{0,3\cdot22,4}{21\%}=32\left(l\right)\)
d) Bảo toàn nguyên tố Lưu huỳnh: \(n_{H_2SO_4\left(lýthuyết\right)}=n_{SO_2\left(thực\right)}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(thực\right)}=0,3\cdot85\%=0,255\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,255\cdot98}{10\%}=249,9\left(g\right)\)
\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)
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) 2KClO3 \(\underrightarrow{to}\) 2KCl + 3O2
b) \(m_{O_2}=62,25-51,65=10,6\left(g\right)\)
\(\Rightarrow n_{O_2}=\dfrac{10,6}{32}=0,33125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,33125\times22,4=7,42\left(l\right)\)
The PT: \(n_{KCl}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\times0,33125=\dfrac{53}{240}\left(mol\right)\)
\(\Rightarrow m_{KCl}=\dfrac{53}{240}\times74,5=16,452\left(g\right)\)