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\(PTHH:4A+3O_2\xrightarrow{t^o} 2A_2O_3\\ \Rightarrow n_{A}=2n_{A_2O_3}\\ \Rightarrow \dfrac{11,2}{M_A}=\dfrac{32}{2M_A+48}\\ \Rightarrow 22,4M_A+537,6=32M_A\\ \Rightarrow 9,6M_A=537,6\\ \Rightarrow M_A=56(g/mol)\)
Vậy A là sắt (Fe)
a, PT: \(4M+3O_2\underrightarrow{t^o}2M_2O_3\)
Ta có: \(n_M=\dfrac{10,8}{M_M}\left(mol\right)\)
\(n_{M_2O_3}=\dfrac{20,4}{2M_M+16.3}\left(mol\right)\)
Theo PT: \(n_M=2n_{M_2O_3}\Rightarrow\dfrac{10,8}{M_M}=2.\dfrac{20,4}{2M_M+16.3}\)
\(\Rightarrow M_M=27\left(g/mol\right)\)
→ M là Nhôm (Al)
b, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\) \(\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
c, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\)
\(n_{HCl}=6n_{Al_2O_3}=1,2\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{1,2}{2}=0,6\left(l\right)\)
d, PT: \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
Theo PT: \(n_{NaOH}=2n_{Al_2O_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{16}{25\%}=64\left(g\right)\)
\(\Rightarrow V_{ddNaOH}=\dfrac{64}{1,25}=51,2\left(ml\right)\)
\(n_{H_2}=\dfrac{0.336}{22.4}=0.015\left(mol\right)\)
\(M+2HCl\rightarrow MCl_2+H_2\)
\(0.015........................0.015\)
\(M_M=\dfrac{0.6}{0.015}=40\left(\dfrac{g}{mol}\right)\)
\(M:Canxi\left(Ca\right)\)
bảo toàn khối lượng ta có
mCl2= 23,4-9,2= 14,2g
nCl2=14,2/71=0,2mol
2A+Cl2-> 2ACl
0,4 0,2
M(A)= 9,2/0,4=23 (Na)
PTHH:2A+Cl2→2ACl
\(m_{Cl_2}=23,4-9,2=14,2\left(g\right)\)
\(n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)
⇒nA=0,2.2=0,4 (mol)
\(M_A=\dfrac{9,2}{0,4}=23\left(g/mol\right)\)
Vậy A là Natri
