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\(n_{Cl_2}=\dfrac{0,224}{22,4}=0,01mol\)
\(2Na+Cl_2\underrightarrow{t^o}2NaCl\)
0,02 0,01 0,02
\(a=m_{Na}=0,02\cdot23=0,46g\)
\(m_{muối}=0,02\cdot58,5=1,17g\)
\(n_{HCl}=\dfrac{58,4.15\%}{36,5}=0,24\left(mol\right)\\ Fe+2HCl\rightarrow\left(t^o\right)FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,24}{2}=0,12\left(mol\right)\\ \Rightarrow V1=V_{H_2\left(đktc\right)}=0,12.22,4=2,688\left(l\right)\\ x=m_{Cu}=m_{hhA}-m_{Fe}=15,68-0,12.56=8,96\left(g\right)\\ b,n_{Cu}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,12+0,14=0,32\left(mol\right)\\ \Rightarrow V2=V_{Cl_2\left(đktc\right)}=0,32.22,4=7,168\left(l\right)\\ y=m_{muối}=m_{AlCl_3}+m_{CuCl_2}=0,12.133,5+0,14.135=34,92\left(g\right)\)
n H2=\(\dfrac{1,12}{22,4}\)=0,05 mol
Zn+2HCl->ZnCl2+H2
0,05---0,1-----0,05---------0,05 mol
ZnO+2HCl->ZnCl2+H2
0,07----0,14---0,07
=m Zn=0,05.65=3,25g
m ZnCl2=0,05.136=6,8g
=>m ZnCl2 pt2 =16,32-6,8=9,52g
=>n ZnCl2=\(\dfrac{9,52}{136}\)=0,07 mol
=>m =3,25+0,07.81=8,92g
=>VHCl=\(\dfrac{0,24}{0,5}\)=0,48l=480ml
PTHH: \(4X+nO_2\underrightarrow{t^o}2X_2O_n\)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
Ta có: \(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\) \(\Rightarrow n_{O_2}=0,025\left(mol\right)\)
\(\Rightarrow n_{X_2O_n}=\dfrac{0,05}{n}\left(mol\right)\) \(\Rightarrow M_{X_2O_n}=\dfrac{3,1}{\dfrac{0,05}{n}}=62n\)
Ta thấy với \(n=1\) \(\Rightarrow M_{X_2O}=62\) \(\Rightarrow M_X=23\)
Vậy kim loại cần tìm là Natri
Bài 1:
\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
PTHH: Fe + 2HCl → FeCl2 + H2
\(m_{H_2}=0,16.2=0,32\left(g\right)\)
\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)
Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)
Bài 12:
Theo ĐLBTKL, ta có:
\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)
PTHH: 2Fe + 3Cl2 --to--> 2FeCl3
0,01-->0,015----->0,01
=> V = 0,015.22,4 = 0,336 (l)
mFeCl3 = 0,01.162,5 = 1,625 (g)