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PT: \(2R+O_2\underrightarrow{t^o}2RO\)
Ta có: \(n_R=\dfrac{13}{M_R}\left(mol\right)\), \(n_{RO}=\dfrac{16,2}{M_R+16}\left(mol\right)\)
Theo PT: \(n_R=n_{RO}\Rightarrow\dfrac{13}{M_R}=\dfrac{16,2}{M_R+16}\Rightarrow M_R=65\left(g/mol\right)\)
→ R là Kẽm (Zn).
PTHH: \(2R+O_2\underrightarrow{t^o}2RO\)
Theo PTHH: \(n_R=n_{RO}\)
\(\Rightarrow\dfrac{3,6}{M_R}=\dfrac{6}{M_R+16}\) \(\Rightarrow M_R=24\)
Vậy kim loại cần tìm là Magie
PT: \(2R+O_2\underrightarrow{t^o}2RO\)
\(n_R=\dfrac{3,6}{M_R}\left(mol\right)\)
\(n_{RO}=\dfrac{6}{M_R+16}\left(mol\right)\)
Theo PT: \(n_R=n_{RO}\Rightarrow\dfrac{3,6}{M_R}=\dfrac{6}{M_R+16}\Rightarrow M_R=24\left(g/mol\right)\)
Vậy: R là Magie.
a) mOxi = moxit - mkim loại = 24 - 19,2 =4,8g
nO2 = 4,8 : 32 = 0,15 mol
VO2 = 0,15.22,4 = 3,36 lít
b) pt: 2R + O2 \(\rightarrow\) 2RO
\(\dfrac{19,2}{R}\) \(\dfrac{24}{R+16}\)
=> \(\dfrac{19,2}{R}=\dfrac{24}{R+16}\)
\(\Leftrightarrow19,2R+307,2=24R\)
\(\Leftrightarrow307,2=4,8R\)
\(\Leftrightarrow R=64\)
Vậy kim loại là Cu
\(n_{A_2O}=\dfrac{9,4}{2M_A+16}\left(mol\right)\)
PTHH: A2O + 2HCl --> 2ACl + H2O
\(\dfrac{9,4}{2M_A+16}\)-->\(\dfrac{9,4}{M_A+8}\)
=> \(\dfrac{9,4}{M_A+8}\left(M_A+35,5\right)=14,9\Rightarrow M_A=39\left(g/mol\right)\)
=> A là K
CTHH: K2O
\(n_M=\dfrac{7,2}{M_M}\left(mol\right)\)
PTHH: M + Cl2 --to--> MCl2
\(\dfrac{7,2}{M_M}\)------------>\(\dfrac{7,2}{M_M}\)
=> \(\dfrac{7,2}{M_M}\left(M_M+71\right)=28,5=>M_M=24\left(Mg\right)\)
Bài 1:
\(n_M=\dfrac{16}{M_M}\left(mol\right)\)
PTHH: 2M + O2 --to--> 2MO
\(\dfrac{16}{M_M}\)---------->\(\dfrac{16}{M_M}\)
=> \(\dfrac{16}{M_M}\left(M_M+16\right)=20\)
=> MM = 64 (g/mol)
=> M là Cu
Bài 2:
\(n_R=\dfrac{16,2}{M_R}\left(mol\right)\)
PTHH: 2R + 3Cl2 --to--> 2RCl3
\(\dfrac{16,2}{M_R}\)------------>\(\dfrac{16,2}{M_R}\)
=> \(\dfrac{16,2}{M_R}\left(M_R+106,5\right)=80,1\)
=> MR = 27 (g/mol)
=> R là Al
1
ADDDLBTKL ta có
\(m_{O_2}=m_{MO}-m_M\\ m_{O_2}=20-16=4g\\ n_{O_2}=\dfrac{4}{32}=0,125\left(mol\right)\\ pthh:2M+O_2\underrightarrow{t^o}2MO\)
0,25 0,125
\(M_M=\dfrac{16}{0,25}=64\left(\dfrac{g}{mol}\right)\)
=> M là Cu
2
ADĐLBTKL ta có
\(m_{Cl_2}=m_{RCl_3}-m_R\\ m_{Cl_2}=80,1-16,2=63,9g\\ n_{Cl_2}=\dfrac{63,9}{71}=0,9\left(mol\right)\\ pthh:2R+3Cl_2\underrightarrow{t^o}2RCl_3\)
0,6 0,9
\(M_R=\dfrac{16,2}{0,6}=27\left(\dfrac{g}{mol}\right)\)
=> R là Al