Đốt cháy 5,4g nhôm trong bình chứa 4,48 lít khí oxi (đktc). Tính khối
lượng oxit thu được sau phản ứng.
mng giúp mình với ạ
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\(a,PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ m_{Al_2O_3}=n.M=0,2.102=20,4\left(g\right)\)
\(b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Lập.tỉ.lệ:\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\Rightarrow Al.dư\\ Theo.PTHH:n_{Al\left(pư\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,2\left(mol\right)\\ n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(pư\right)}=0,4-0,2=0,2\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=n.M=0,1=102=10,2\left(g\right)\)
a) PTHH: 4Al + 3O2 =(nhiệt)=> 2Al2O3
nAl = \(\frac{5,4}{27}=0,2\left(mol\right)\)
b) nO2 = \(\frac{0,2\times3}{4}=0,15\left(mol\right)\)
=> VO2(đktc) = 0,15 x 22,4 = 3,36 lít
c) nAl2O3 = \(\frac{0,2\times2}{4}=0,1\left(mol\right)\)
=> mAl2O3 = 0,1 x 102 = 10,2 gam
nO2 = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
pt: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Theo pt: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
=> nKMnO4 thực tế = 0,6:\(\dfrac{90}{100}=\dfrac{2}{3}\left(mol\right)\)
mKMnO4 = \(\dfrac{2}{3}.158=\dfrac{316}{3}g\)
\(n_{Al}=\dfrac{3,24}{27}=0,12mol\)
a)\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) \(\Rightarrow\) phản ứng hóa hợp.
b)0,12 0,09 0,06
\(m_{Al_2O_3}=0,06\cdot102=6,12g\)
c)\(V_{O_2}=0,09\cdot22,4=2,016l\)
Ta có: n\(O_2\)=\(\dfrac{6.72}{22.4}\)=0.3 (mol)
PTHH: 4Al + 3O2 ______> 2Al2O3 (1)
Ta có: theo (1): nAl =\(\dfrac{4}{3}n_{O_2}\)=\(\dfrac{4}{3}0.3=0.4\left(mol\right)\)
=> mAl = 0.4 . 27=10.8(g)
PTHH: 4Al+3O2->to 2Al2O3
4 3 2 (mol)
0,3 (mol)
nO2= V/22,4=6,72/22,4=0,3 (mol)
nAl= nO2.4/3=0,3.4/3=0.4 (mol)
mAl=n.M=0,4.27=10,8 (g)
a.\(n_{CH_4}=\dfrac{V_{CH_4}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,3 0,6 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,6.22,4=13,44l\)
b.
\(n_P=\dfrac{m_P}{M_P}=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
0,1 0,05 ( mol )
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,05.142=7,1g\)
\(n_{KMnO_4}=\dfrac{18.96}{158}=0.12\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.12...........................................0.06\)
\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.08.....0.06.......0.04\)
\(m_{Al\left(dư\right)}=\left(0.2-0.08\right)\cdot27=3.24\left(g\right)\)
\(m_{Al_2O_3}=0.04\cdot102=4.08\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
0,2 0,1
\(m_{Al_2O_3}=0,1\cdot102=10,2g\)
\(n_{Al}=\dfrac{5.4}{27}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)
PTHH : 4Al + 3O2 ---to---> 2Al2O3
0,2 0,1
Ta thấy : \(\dfrac{0.2}{4}< \dfrac{0.2}{3}\) => Al đủ , O2 dư
\(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)