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\(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(2X+O_2\underrightarrow{t^o}2XO\)
\(\dfrac{13}{X}\) 0,1
\(\Rightarrow\dfrac{13}{X}=0,1\cdot2\Rightarrow X=65\)
Vậy X là kẽm Zn.
\(m_{ZnO}=0,2\cdot81=1,62g\)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2R + O2 --to--> 2RO
0,2 0,.1
=> \(M_R=\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\)
=> R: Zn
\(Đặt.kim.loại.kiềm:A\\ 2A+2HCl\rightarrow2ACl+H_2\\ m_{muối}-m_{kl}=m_{Cl^-}\\ \Leftrightarrow m_{Cl^-}=7,45-3,9=3,55\left(g\right)\\ \Rightarrow n_{HCl}=n_{Cl^-}=\dfrac{3,55}{35,5}=0,1\left(mol\right)\\ \Rightarrow n_A=n_{ACl}=n_{HCl}=0,1\left(mol\right)\\ a,M_A=\dfrac{3,9}{0,1}=39\left(\dfrac{g}{mol}\right)\\ \Rightarrow A\left(I\right):Kali\left(K=39\right)\\ b,n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ c,m_{ddHCl}=\dfrac{0,1.36,5.100}{31,7}=\dfrac{3650}{317}\left(g\right)\\ \Rightarrow V_{ddHCl}=\dfrac{\dfrac{3650}{317}}{1,15}\approx10,012\left(g\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH : \(2X+O_2\underrightarrow{t^o}2XO\)
0,3 0,15 /mol
Ta có : \(0,3=\dfrac{19,2}{X}\Rightarrow X=64\) => X là Cu
\(m_{CuO}=80.0,3=24\left(g\right)\)
Gọi R là kim loại cần tìm.
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(2R+O_2\underrightarrow{t^o}2RO\)
\(\dfrac{19,2}{R}\) 0,15
\(\Rightarrow\dfrac{19,2}{R}=0,15\cdot2\Rightarrow R=64\Rightarrow Cu\)
Khối lượng oxit: \(m_{CuO}=0,3\cdot80=24g\)
\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
\(2R+2nHCl\rightarrow2RCl_n+nH_2\)
\(\dfrac{1.2}{n}......1.2...............0.6\)
\(M_R=\dfrac{14.4}{\dfrac{1.2}{n}}=12n\)
\(BL:n=2\Rightarrow R=24\)
\(R:Mg\)
\(m_{MgCl_2}=0.6\cdot95=57\left(g\right)\)
\(m_{dd}=14.4+146-0.6\cdot2=159.2\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{57}{159.2}\cdot100\%=35.8\%\)
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