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Sodium reacts violently in chlorine gas to form sodium chloride. If [tex]2.5 \, \text{mol Na}[/tex] and [tex]1.1 \, \text{mol Cl}_2[/tex] are placed in a flask and allowed to react, how much sodium chloride could be formed?

[tex]2 \, \text{Na} + \text{Cl}_2 \rightarrow 2 \, \text{NaCl}[/tex]

A. 3.6 mol, because the moles of reactants must equal the moles of products

B. 2.5 mol, because the amount of product is determined by the limiting reactant, sodium

C. 2.2 mol, because the amount of product is determined by the limiting reactant, chlorine

D. 2.4 mol, because 2 mol NaCl form for each 3 mol of reactants that react

Sagot :

GinnyAnswer
To determine how much sodium chloride ([tex]\( \text{NaCl} \)[/tex]) could be formed when [tex]\( 2.5 \, \text{mol} \, \text{Na} \)[/tex] and [tex]\( 1.1 \, \text{mol} \, \text{Cl}_2 \)[/tex] are placed in a flask and allowed to react, we start by examining the reaction stoichiometry given by the balanced chemical equation:

[tex]\[ 2 \, \text{Na} + \text{Cl}_2 \rightarrow 2 \, \text{NaCl} \][/tex]

This equation tells us that 2 moles of sodium (Na) react with 1 mole of chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]) to produce 2 moles of sodium chloride ([tex]\( \text{NaCl} \)[/tex]).

### Step 1: Identify the Limiting Reactant
- Since 2 moles of Na react with 1 mole of [tex]\( \text{Cl}_2 \)[/tex], the mole ratio of Na to [tex]\( \text{Cl}_2 \)[/tex] is 2:1.

Let's determine how many moles of [tex]\( \text{NaCl} \)[/tex] can be formed by each reactant separately:
1. Sodium (Na):
- We have 2.5 moles of Na.
- According to the stoichiometry, 2 moles of Na produce 2 moles of [tex]\( \text{NaCl} \)[/tex].
- Therefore, [tex]\( 2.5 \)[/tex] moles of Na would produce [tex]\( 2.5 \)[/tex] moles of [tex]\( \text{NaCl} \)[/tex],
assuming there is enough [tex]\( \text{Cl}_2 \)[/tex].

2. Chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]):
- We have 1.1 moles of [tex]\( \text{Cl}_2 \)[/tex].
- According to the stoichiometry, 1 mole of [tex]\( \text{Cl}_2 \)[/tex] produces 2 moles of [tex]\( \text{NaCl} \)[/tex].
- Therefore, [tex]\( 1.1 \)[/tex] moles of [tex]\( \text{Cl}_2 \)[/tex] would produce [tex]\( 2 \times 1.1 = 2.2 \)[/tex] moles of [tex]\( \text{NaCl} \)[/tex],
assuming there is enough Na.

### Step 2: Determine the Amount of NaCl Formed by the Limiting Reactant
- The reactant that produces fewer moles of [tex]\( \text{NaCl} \)[/tex] is the limiting reactant.
- Since chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]) produces only [tex]\( 2.2 \)[/tex] moles of [tex]\( \text{NaCl} \)[/tex] and still follows the stoichiometric ratio, [tex]\( \text{Cl}_2 \)[/tex] is the limiting reactant.

Hence, the amount of [tex]\( \text{NaCl} \)[/tex] that could be formed is determined by the limiting reactant, which is chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]) in this case.

### Conclusion
- The maximum amount of sodium chloride ([tex]\( \text{NaCl} \)[/tex]) that can be formed is [tex]\( 2.2 \)[/tex] moles.

Thus, the correct answer is:
C. 2.2 mol, because the amount of product is determined by the limiting reactant, chlorine.
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