For the following reaction, 15.4 grams of chlorine gas are allowed to react with 49.6 grams of sodium iodide. chlorine (g) sodium iodide (s) sodium chloride (s) iodine (s) What is the maximum amount of sodium chloride that can be formed

19.3 g of NaCl can be produced

Explanation:

We state the reaction:

Cl₂ (g) + 2NaI (s) →  2NaCl (s) + I₂ (s)

We need to determine limiting reagent:

15.4 g . 1mol /70.9g = 0.217 moles of chlorine

49.6 g . 1mol / 149.89g = 0.331 moles of NaI

Ratio is 1:2. 1 mol of chlorine reacts to 2 moles of NaI

0.217 moles may react to (0.217 . 2)/1 = 0.434 moles of NaI

It is ok to say the NaI is the limting reactant because we need 0.434 moles of it and we only have 0.331.

Ratio is 2:2.

0.331 moles of NaI can produce 0.331 moles of NaCl

We convert mass to moles: 0.331 mol . 58.45g /mol = 19.3 g