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viva [34]
3 years ago
13

Limiting Reactant

Chemistry
1 answer:
liraira [26]3 years ago
3 0

Answer:

(1) Cl₂ is the limiting reactant.

(2) 8.18 g

Explanation:

  • 2Na(s) + Cl₂(g) → 2NaCl(s)

First we <u>convert the given masses of reactants into moles</u>, using their <em>respective molar masses</em>:

  • 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 <u>calculate how many moles of NaCl are formed</u>, <em>using the limiting reactant</em>:

  • 0.070 mol Cl₂ * \frac{2molNaCl}{1molCl_2} = 0.14 mol NaCl

Finally we <u>convert NaCl moles into grams</u>:

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