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nP = 6,2/31 = 0,2 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
Mol: 0,2 ---> 0,25
4R + nO2 -> (t°) 2R2On
Mol: 1/n <--- 0,25
M(R) = 32(1/n) = 32n (g/mol)
Xét:
n = 1 => Loại
n = 2 => R = 64 => R là Cu
n = 3 => Loại
Vậy R là Cu
2Al+3H2SO4->Al2(SO4)3+3H2
0,2-----------------------------------0,3
n Al=0,2 mol
=>VH2=0,3.22,4=6,72l
b)
XO+H2-to>X+H2O
0,3-------------0,3
=>0,3=\(\dfrac{19,5}{X}\)
=>X là Zn( kẽm)
a.\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,3 ( mol )
\(V_{H_2}=0,3.22,4=6,72l\)
b.\(n_X=\dfrac{19,5}{M_X}\)
\(XO+H_2\rightarrow\left(t^o\right)X+H_2O\)
\(\dfrac{19,5}{M_X}\) \(\dfrac{19,5}{M_X}\) ( mol )
Ta có:
\(\dfrac{19,5}{M_X}=0,3\)
\(\Leftrightarrow M_X=65\)
=> X là kẽm (Zn)
\(a,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:2Al+3H_2SO_{4\left(loãng\right)}\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Theo.pt:n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ b,PTHH:RO+H_2\underrightarrow{t^o}R+H_2O\\ Mol:0,6\leftarrow0,6\rightarrow0,6\\ M_R=\dfrac{38,4}{0,6}=64\left(\dfrac{g}{mol}\right)\\ \Rightarrow R.là.Cu\)
nKMnO4=94,8:158=0,6(mol)
PTHH: 2KMnO4-t--> K2MnO4+MnO2+O2
0,6----------------------------------->0,3(mol)
=>V= VO2=0,3. 22,4= 6,72(l)
b ) 40%nO2 =40%.0,3=0,12(mol)
2R + O2 -t--->2RO
0,24(mol)<- 0,12
=> M(Khối lượng Mol ) R= m:n=5,76:0,24=24(G/MOL)
=> R là Mg
a)-\(n_{KMnO_4}=\dfrac{m}{M}=\dfrac{94,8}{158}=0,6\left(mol\right)\)
-PTHH: \(2KMnO_4\rightarrow^{t^0}K_2MnO_4+MnO_2+O_2\uparrow\)
2 1
0,6 0,3
\(\Rightarrow V_{O_2\left(đktc\right)}=n.22,4=0,3.22,4=6,72\left(l\right)\)
b)-\(V_{O_2\left(cd\right)}=6,72.\dfrac{40}{100}=2,688\left(l\right)\)
\(\Rightarrow n_{O_2}=\dfrac{V}{22,4}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
-PTHH: \(2R+O_2\rightarrow^{t^0}2RO\)
2 1
0,24 0,12
\(m_R=n.M=5,76\left(g\right)\)
\(\Rightarrow0,24.M_R=5,76\)
\(\Rightarrow M_R=24\) (g/mol)
-Vậy R là Crom
\(a.PTHH:2B+O_2\overset{t^o}{--->}2BO\left(1\right)\)
b. Áp dụng ĐLBTKL, ta có:
\(m_B+m_{O_2}=m_{BO}\)
\(\Leftrightarrow m_{O_2}=8-4,8=3,2\left(g\right)\)
c. Ta có: \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(lít\right)\)
Mà: \(V_{O_2}=\dfrac{1}{5}.V_{kk}\)
\(\Leftrightarrow V_{kk}=2,24.5=11,2\left(lít\right)\)
d. Theo PT(1): \(n_B=2.n_{O_2}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow M_B=\dfrac{4,8}{0,2}=24\left(\dfrac{g}{mol}\right)\)
Vậy B là magie (Mg)
\(e.PTHH:2xB+yO_2\overset{t^o}{--->}2B_xO_y\left(2\right)\)
Theo PT(2): \(n_B=\dfrac{2x}{y}.n_{O_2}=\dfrac{2x}{y}.0,1=\dfrac{0,2x}{y}\left(mol\right)\)
\(\Rightarrow M_B=\dfrac{4,8}{\dfrac{0,2x}{y}}=\dfrac{4,8y}{0,2x}=12.\dfrac{2y}{x}\left(mol\right)\)
Biện luận:
| 2y/x | 1 | 2 | 3 |
| MB | 12 | 24 | 36 |
| loại | Mg | loại |
Vậy B là kim loại magie (Mg)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{CuO}=n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right);n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ V_{kk\left(đktc\right)}=\dfrac{100.1,12}{20}=5,6\left(l\right)\\ b,m_{CuO}=0,1.80=8\left(g\right)\\ c,2R+O_2\rightarrow\left(t^o\right)2RO\\ n_R=2.n_{O_2}=2.0,05=0,1\left(mol\right)\\ M_R=\dfrac{2,4}{0,1}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow R:Magie\left(Mg=24\right)\)
a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
Theo PTHH : $n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)$
$V_{O_2} = 0,15.22,4 = 3,36(lít)$
b) $2 KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,1(mol)$
$m_{KClO_3} = 0,1.122,5 = 12,25(gam)$
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2-^{t^o}>2Al_2O_3\)
tỉ lệ: 4 : 3 : 2
n(mol) 0,2---->0,15---->0,1
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)
tỉ lệ: 2 : 2 : 3
n(mol) 0,1<-------------------------0,15
\(m_{KClO_3}=n\cdot M=0,1\cdot\left(39+35,5+16\cdot3\right)=12,25\left(g\right)\)
4Al+3O2-to>2Al2O3
0,04---0,03------0,02 mol
n Al=\(\dfrac{1,08}{27}\)=0,04 mol
=>VO2=0,03.22,4=0,672l
b)
2A+O2-to>2AO
0,06--0,03 mol
=>\(\dfrac{3,84}{A}=0,06\)
=>A=64 :=>Al là Đồng
Al là đồng là sao ạ =)))