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\(a,PTHH:4A+3O_2\underrightarrow{t^o}2A_2O_3\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_A+m_{O_2}=m_{A_2O_3}\\ \Rightarrow m_{O_2}=m_{A_2O_3}-m_A=20,4-10,8=9,6\left(g\right)\)
\(\Rightarrow n_{O_2}=\dfrac{m}{M}=\dfrac{9,6}{32}=0,3\left(mol\right)\\ Theo.PTHH:n_A=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,3=0,4\left(mol\right)\\ \Rightarrow M_A=\dfrac{m}{n}=\dfrac{10,8}{0,4}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A.là.Al\left(nhôm\right)\)
\(b,V_{O_2\left(đktc\right)}=n.22,4=0,4.22,4=8,96\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=V_{O_2\left(đktc\right)}.5=8,96.5=44,8\left(l\right)\)
\(a,4A+3O_2\rightarrow\left(t^o\right)2A_2O_3\\ Theo.ĐLBTKL:\\ m_A+m_{O_2}=m_{A_2O_3}\\ \Leftrightarrow10,8+m_{O_2}=20,4\\ \Leftrightarrow m_{O_2}=9,6\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\ n_A=\dfrac{4}{3}.0,3=0,4\left(mol\right)\Rightarrow M_A=\dfrac{m_A}{n_A}=\dfrac{10,8}{0,4}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Nhôm\left(Al=27\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.V_{O_2\left(đktc\right)}=5.\left(0,3.22,4\right)=33,6\left(l\right)\)
a) PTHH: 4P+5O2-----to---> 2P2O5
0,2 0,25 0,1
b)\(n_{P_2O_5}=\dfrac{m}{M}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
\(m_P=n.M=0,2.31=6,2\left(gam\right)\)
c) \(V_{O_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
a) nAl=0,2(mol)
PTHH: 4Al +3 O2 -to-> 2 Al2O3
nO2=3/4. 0,2=0,15(mol)
=>V(O2,đktc)=0,15.22,4=3,36(l)
b) V(kk,đktc)=3,36.5=16,8(l)
a: \(4Al+3O_2\rightarrow2Al_2O_3\)
b: \(n_{Al}=\dfrac{21.6}{27}=0.8\left(mol\right)\)
\(\Leftrightarrow n_{Al_2O_3}=0.4\left(mol\right)\)
\(m_{Al_2O_3}=0.4\cdot102=40.8\left(g\right)\)
c: \(n_{O_2}=0.6\left(mol\right)\)
\(V_{O_2}=0.6\cdot22.4=13.44\left(lít\right)\)
a) 4Al + 3O2 --to--> 2Al2O3
b) \(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
0,8-->0,6-------->0,4
=> \(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)
c) \(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
d) \(V_{kk}=13,44:20\%=67,2\left(l\right)\)
2Cu+O2-to>2CuO
0,4-----0,2-----------0,4 mol
n Cu=\(\dfrac{12,8}{64}\)=0,4 mol
=>m CuO=0,4.56=22,4g
=>Vkk=0,2.22,4.5=22,4l
nAl = 2,7/27 = 0,1 (mol)
PTHH: 4Al + 3O2 -> (t°) 2Al2O3
Mol: 0,1 ---> 0,075 ---> 0,05
mAl2O3 = 0,05 . 102 = 5,1 (g)
VO2 = 0,075 . 22,4 = 1,68 (l)
Vkk = 1,68 . 5 = 8,4 (l)
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,1 0,075 0,05 ( mol )
\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,05.102=5,1g\)
\(V_{kk}=V_{O_2}.5=\left(0,075.22,4\right).5=8,4l\)
a) C + O2 --to--> CO2
b) \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)
PTHH: C + O2 --to--> CO2
_____0,2->0,2------>0,2
=> mCO2 = 0,2.44 = 8,8 (g)
c) VO2 = 0,2.22,4 = 4,48(l)
=> Vkk = 4,48.5 = 22,4 (l)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
a)\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,4 0,2 0,4
\(V_{O_2}=0,2\cdot22,4=4,48l\)
\(V_{kk}=5V_{O_2}=5\cdot4,48=22,4l\)
b)\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,4
\(m_{H_2O}=0,4\cdot18=7,2g\)
\(n_{CO_2}=\dfrac{52,8}{44}=1,2\left(mol\right);n_{H_2O}=\dfrac{10,8}{18}=0,6\left(mol\right)\)
\(PTHH:2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Mol: 0,6 1,5 1,2 0,6
\(\Rightarrow a=m_{C_2H_2}=0,6.26=15,6\left(g\right)\)
Ta có:\(V_{kk}=5.V_{O_2}=5.1,5.22,4=168\left(l\right)\)