with regard to the reaction of sodium with chlorine (2 Na + Cl2 = 2 NaCl); if 1.0 mole of Cl2 is heated in the presence of 6.0 moles of sodium metal, how many moles of NaCl will be formed?

14.0

14.0 mole of 2NaCl

To determine the number of moles of NaCl that will be formed in the reaction of sodium with chlorine, first, we need to identify the limiting reactant.

The stoichiometry of the reaction tells us that 2 moles of sodium react with 1 mole of chlorine to produce 2 moles of NaCl. Therefore, the balanced equation indicates that the reaction ratio is 2:1:2 (Na:Cl2:NaCl).

Given that 1.0 mole of Cl2 is present, and assuming the reaction is complete, we need to compare this with the amount of sodium present. In this case, we have 6.0 moles of sodium.

To find the limiting reactant, we need to calculate the moles of NaCl that would be formed if all of the Cl2 reacted:

1 mole of Cl2 will produce 2 moles of NaCl.

So, 1.0 mole of Cl2 will produce (2 moles NaCl/1 mole Cl2) = 2.0 moles of NaCl.

Next, we compare this with the amount of Na available:

6.0 moles of Na would produce (2 moles NaCl/2 moles Na) = 6.0 moles of NaCl.

Since we have less NaCl (2.0 moles) formed from the Cl2 compared to the Na (6.0 moles), the Cl2 is the limiting reactant.

Therefore, when 1.0 mole of Cl2 is heated in the presence of 6.0 moles of sodium metal, the amount of NaCl formed is 2.0 moles.