\(R+\dfrac{1}{2}O_2\rightarrow\left(t^o\right)RO\)
\(n_{RO}=\dfrac{6}{M_R+16}\)
\(R+\dfrac{1}{2}O_2\rightarrow\left(t^o\right)RO\)
\(\dfrac{6}{M_R+16}\) <---- \(\dfrac{6}{M_R+16}\) ( mol )
Ta có:
\(\dfrac{6}{M_R+16}.M_R=3,6\)
\(\Leftrightarrow6M_R=3,6M_R+57,6\)
\(\Leftrightarrow M_R=24\) ( g/mol )
=> R là Magie (Mg)
Áp dụng đlbtkl, ta có:
mR + mO2 = mR2On
=> mO2 = 6 - 3,6 = 2,4 (g)
nO2 = \(\dfrac{2,4}{32}=0,075\left(mol\right)\)
PTHH: 4R + nO2 ---to---> 2R2On
\(\dfrac{0,3}{n}\) 0,075
\(M_R=\dfrac{3,6}{\dfrac{0,3}{n}}=12n\left(\dfrac{g}{mol}\right)\)
Xét:
n = 1 => Loại
n = 2 => R = 24 => Mg
n = 3 => Loại
Vậy R là Mg
