The unbalanced equation below shows the combustion of methane.
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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 *
= 0.14 mol NaCl
Finally we <u>convert NaCl moles into grams</u>:
- 0.14 mol NaCl * 58.44 g/mol = 8.18 g
Answer:
You can find in the given attachemnet
