X + O2 \(\underrightarrow{to}\) XO2
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{XO_2}=n_{O_2}=0,1\left(mol\right)\)
\(\Rightarrow M_{XO_2}=\dfrac{4,4}{0,1}=44\left(g\right)\)
\(\Leftrightarrow M_X+32=44\)
\(\Leftrightarrow M_X=12\left(g\right)\)
Vậy X là cacbon C
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH:
\(X+O_2\underrightarrow{t^o}XO_2\)
0,1-->0,1--->0,1(mol)
\(M_{XO_2}=\dfrac{m}{n}=\dfrac{4,4}{0,1}=44\left(g/mol\right)\)
Ta có: \(M_{XO_2}=M_X+M_{O_2}\)
\(\Leftrightarrow44=M_X=32\)
\(\Rightarrow M_X=12\left(g/mol\right)\)
Vậy X là Cacbon (C)
