a,
Giả sử có 3g Mg, 4g S
\(\Rightarrow n_{Mg}=0,125\left(mol\right);n_S=0,125\left(mol\right)\)
\(n_{Mg}:n_S=0,125:0,125=1:1\)
Vậy CTHH là MgS
b, \(n_{Mg}=\frac{1}{3}\left(mol\right);n_S=0,25\left(mol\right)\)
\(Mg+S\underrightarrow{^{to}}MgS\)
Tạo 0,25 mol MgS. Dư 1/12 mol Mg
\(m_{MgS}=14\left(g\right)\)
\(m_{Mg_{dư}}=2\left(g\right)\)