Answer:
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
