Answer
(a) 2Na(s) + I$_{2}$(s) → 2NaI (Sodium Iodide)
(b) Sr(s) + H$_{2}$(g) →H$_{2}$Sr (Strontium Dihydride)
(c) Ba(s) + Cl$_{2}$(g) → BaCl$_{2}$ (Barium Chloride)
(d) Mg(s) + O$_{2}$(g) →MgO$_{2}$ (Magnesium Oxide)
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