Question

What is the greatest amount of BaCl2 (in grams) that can be made with 114g of ba and 186g of cl2?

Answer 1

Answer:

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.