the synthesis of sodium chloride reaction how many grams of sodium chloride could be produced from 100.0 g of sodium and 12.5 L of chlorine ( at STP ) ? Be sure to identify the limiting reactant .

Answer 1

In any question of this sort, you must first write the correctly balanced equation for the reaction:

2Na + Cl₂ ==> 2NaCl

Next, we must identify the limiting reactant. An easy way to do this is to divided the MOLES of each reactant by the corresponding coefficient in the balanced equation.

For Na: 100.0 g Na x 1 mol Na / 23.0 g = 4.35 mols Na (÷2->2.17)

For Cl₂: 12.5 L Cl₂ x 1 mol Cl2 / 22.4 L = 0.558 moles Cl₂ (÷1->0.56) Recall 1 mol gas @STP = 22.4 L

Since 0.56 is less than 2.17, Cl₂ is the limiting reactant.

Now, we use the MOLES of the limiting reactant to find the theoretical yield of the product.

0.558 mols Cl₂ x 2 mols NaCl / mol Cl₂ x 58.4 g NaCl / mol = 65.2 g g NaCl (3 sig. figs.)