\(n_{Na_2O}=\dfrac{124}{62}=2\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
2------->2
\(\rightarrow n_{CO_2}=\dfrac{2.1}{2}=1\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}m_{H_2O}=2.18=36\left(g\right)\\m_{CO_2}=1.44=44\left(g\right)\end{matrix}\right.\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{O_2}=0,2.32=6,4\left(g\right)\)
Bảo toàn khối lượng:
mX + mO2 = mCO2 + mH2O
\(\Rightarrow m_1=m_X=36+44-6,4=73,6\left(g\right)\)