5.6 mols of sodium reacts with 3.4 mols of chlorine. How many mols of sodium chloride are formed? What is the limiting reactant?

Question

Caleb Adams · Accepted Answer

These questions are always problematic. If we assume standard conditions, NATRIUM is the limiting reagent.

#Na(s) + 1/2Cl_2(g) rarr NaCl(s)# We assume standard conditions. Thus, there are 5.6 mol sodium metal and 3.4 mols #Cl_2# gas, that is 6.8 mol chlorine atoms . Thus the metal is the reagent in deficiency, and we can form 5.6 mol of salt. Note that I can only think of one group of gases that are unimolecular under standard conditions. Which are they? The majority of elemental gases are BIMOLECULAR under standard conditions, and this is something that an examiner would expect you to know.

Wyatt Adkins · Answer

undefined To find the number of moles of sodium chloride formed, we need to determine the limiting reactant. First, we need to compare the stoichiometric coefficients of sodium and chlorine in the balanced equation to see which one is the limiting reactant. The balanced equation for the reaction between sodium and chlorine is:

2 Na + Cl2 -> 2 NaCl

Given:
Moles of sodium (Na) = 5.6 moles
Moles of chlorine (Cl2) = 3.4 moles

Using stoichiometry, we can calculate the moles of sodium chloride formed:

Moles of sodium chloride = (moles of limiting reactant) * (moles of product per mole of reactant)

Now, let's calculate the moles of sodium chloride formed using both reactants:

For sodium:
Moles of sodium chloride = (5.6 moles Na) * (2 moles NaCl / 2 moles Na) = 5.6 moles NaCl

For chlorine:
Moles of sodium chloride = (3.4 moles Cl2) * (2 moles NaCl / 1 mole Cl2) = 6.8 moles NaCl

Since sodium chloride cannot be formed in fractional moles, the limiting reactant is sodium. Therefore, 5.6 moles of sodium chloride are formed, and sodium is the limiting reactant.