\(n_{H_2}=\dfrac{1,2}{2}=0,6\left(mol\right)\)
so phtu có trong 1,2g H2:
0,6.6.1023 = 3,6.1023
gọi x la so g Mg
\(n_{Mg}=\dfrac{x}{24}\left(mol\right)\)
so phtu Mg: \(\dfrac{x}{24}.6.10^{23}=3,6.10^{23}\)
\(\Rightarrow6x.10^{23}=86,4.10^{23}\)
\(\Rightarrow x=14,4g\)