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\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)
a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 0,6 0,2 0,6
b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
=>\(n_{Al}=0.4\left(mol\right)\)
\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)
c: \(4Al+3O_2\rightarrow2Al_2O_3\)
0,4 0,2
\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 4Al + 3O2 ---to→ 2Al2O3
Mol: 0,2 0,15 0,1
b) \(V_{O_2}=0,15.24,79=3,7185\left(mol\right)\)
c) \(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ a,4Al+3O_2\underrightarrow{^{to}}2Al_2O_3\\ 0,2.......0,15........0,1\left(mol\right)\\ b,V_{O_2\left(25^oC,1bar\right)}=24,79.0,15=3,7185\left(l\right)\\ c,m_A=m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
\(n_{Mg}=\dfrac{m}{M}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)-0,2--0,1--0,2\\ V_{O_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)
Ta có: \(n_{CuSO_4}=0,3\left(mol\right)\)
a, PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
______0,2____0,3_________________0,3 (mol)
b, \(m_{Al}=0,2.27=5,4\left(g\right)\)
c, \(m_{Cu}=0,3.64=19,2\left(g\right)\)
Bạn tham khảo nhé!
a, \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
b, \(n_{CH_3COOH}=\dfrac{3,6}{60}=0,06\left(mol\right)\)
Theo PT: \(n_{\left(CH_3COO\right)_2Mg}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,03\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,03.142=4,26\left(g\right)\)
\(V_{H_2}=0,06.22,4=1,344\left(l\right)\)
a, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,2\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4}{6,72}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)
b, Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=0,7\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(l\right)\)
nAL=54:27=2 mol
nAL2O3=1mol
PTHH: 4Al+3O2=>2Al2O3
2 1
2->3/2<==3/2
=> mO2=1,5.32=48g
=> V O2=1,5.22,4=33,6l
làm sao bạn tính ra được số mol của nhôm oxit là 1 vậy