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CHÚC BẠN HỌC TỐT!
Ta có: \(m_{Al2O3}=326,4.40\%=130,56\left(g\right)\)
\(\Rightarrow n_{Al2O3}=\dfrac{130,56}{27.2+3.16}=1,28\left(mol\right)\)
PTHH: \(2Al_2O_3\xrightarrow[criolit]{đpnc}4Al+3O_2\)
pư...........1,28...............2,56.......1,92 (mol)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al\left(LT\right)}=27.2,56=69,12\left(g\right)\\V_{O2}=22,4.1,92=43,008\left(l\right)\end{matrix}\right.\)
\(\Rightarrow m_{Al\left(TT\right)}=69,12.90\%=62,208\left(g\right)\)
Vậy...........
\(a) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{Al} = \dfrac{5,4}{27} = 0,2(mol)\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,1(mol) \Rightarrow m_{Al_2O_3} = 0,1.102 = 10,2(gam)\\ b) n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3(mol) \Rightarrow m_{KMnO_4} = 0,3.158 = 47,4(gam)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,2.158=47,4\left(g\right)\)
Bạn tham khảo nhé!
doi 280kg = 280000g
\(m_{tc}=\dfrac{25.280000}{100}=70000g\)
\(m_{CaCO_3}=280000-70000=210000g\)
\(n_{CaCO_3}=\dfrac{210000}{100}=2100\left(mol\right)\)
CaCO3 \(\underrightarrow{t^o}\) CaO + CO2
de: 2100 \(\rightarrow\) 2100 \(\rightarrow\) 2100 (mol)
\(m_{CaO}=2100.56=117600g\)
\(V_{CO_2}=2100.22,4=47040l\)
2Al2O3 -> 4Al + 3O2
204 108 67,2
13,56 x y
x= 13,56 × 108 : 204 = 69,12g
y= 69,12 × 67,2 : 108 = 43,008g
Hx= 69,12 × 90 : 100 = 62,208
Hy= 43,008 × 90 : 100 = 38,70
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\a, PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ b,n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3.0,2}{4}=0,15\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=2.n_{O_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow m_{KMnO_4}=158.0,3=47,4\left(g\right)\)
a/ nAl= 54/27= 2(mol)
nO2=48/32=1,5(mol)
PTHH: 2 Al2O3 -to-> 4 Al +3 O2
Ta có: 2/4 = 1,5/3
=> P.ứ hết
=> nAl2O3= 1/2. nAl=1/2. 2=1(mol)
=> mAl2O3=1.102=102(g)
b) %mAl2O3= (102/127,5).100= 80%
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,4}{4}< \dfrac{0,4}{3}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=0,1.22,4=2,24\left(l\right)\)
Lập phương trình hóa học:
Al+O2---->Al2O3
4Al+3O2---->2AlO3
Áp dụng đinh luật bảo toàn khối lượng ta có:
mAl + mO2=mAl2O3
=>mO2=mAl2O3 - mAl
=>mO2=20,4 - 10,8=9,6(g)
Số mol của 9,6g khí oxi là:
ADCT: n=m\M=>nO2=9,6\32=>nO2=0,3(mol)
n=V\22,4=>VO2=nO2 . 22,4=0,3 . 22,4=6,72(l)
CHÚC BẠN HỌC TỐT!
Ta có: m\(m_{Al2O3}=326,4.40\%=130,56\left(g\right)\)
\(\Rightarrow n_{Al2O3}=\dfrac{130,56}{2.27+3.16}=1,28\left(mol\right)\)
PTHH: \(3Al_2O_3\xrightarrow[criolit]{đpnc}4Al+3O_2\)
pư.............1,28.............2,56.....1,92 (mol)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=27.2,56=69,12\left(g\right)\\V_{O2}=22,4.1,92=43,008\left(l\right)\end{matrix}\right.\)
Vậy.......