Question

Limiting Reactant

12.0 grams of sodium reacts with 5.00 grams of chlorine. What mass of sodium
chloride could be produced?
Nas) +
Cl2(g) →
NaCls)
(1)
(2)
Identify the limiting reactant.
Determine the amount of sodium chloride produced.

Answer 1

(1) Cl₂ is the limiting reactant.

(2) 8.18 g

Step-by-step explanation:

• 2Na(s) + Cl₂(g) → 2NaCl(s)

First we convert the given masses of reactants into moles, using their respective molar masses:

• Na ⇒ 12.0 g ÷ 23 g/mol = 0.522 mol Na

• Cl₂ ⇒ 5.00 g ÷ 70.9 g/mol = 0.070 mol Cl₂

0.070 moles of Cl₂ would react completely with (2 * 0.070) 0.14 moles of Na. There are more Na moles than that, so Na is the reactant in excess while Cl₂ is the limiting reactant.

Then we calculate how many moles of NaCl are formed, using the limiting reactant:

• 0.070 mol Cl₂ *
  `(2molNaCl)/(1molCl_2)` = 0.14 mol NaCl

Finally we convert NaCl moles into grams:

• 0.14 mol NaCl * 58.44 g/mol = 8.18 g