What is the greatest amount of bacl2 (in grams) that can be made with 114g of ba and 186g of cl2?
what is the limiting reactant?
which reactant is in excess and how? and how many grams are left?

Explanation:

The equation for this chemical reaction is:

Ba + Cl2 -> BaCl2

So 1 moles of Ba react with 2 mole of Cl2 to form 1 mole of BaCl2.

Given 114g of Ba and 186g of Cl2 and molar mass of 137g for Ba and 71g for Cl2,

there are 114/137 = 0.83 mole of Ba and 186/71 = 2.62 mole of Cl2.

Ba is the limiting reactant as it will be used up.

Cl2 is the reactant in excess and there will be 2.62 - 0.83*2 = 0.96 mole or 0.96*71 = 68.16g left.

Moles of BaCl2 formed = moles of Ba = 0.83

As mass of 1 mole of BaCl2 is 208.23g, 0.83 mole or 0.83 * 208.23 = 172.83g of BaCl2 can be made.