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How many moles of chlorine could be produced by decomposing [tex]$157 \, g \, NaCl$[/tex]?

[tex]2 \, NaCl \rightarrow 2 \, Na + Cl_2[/tex]

A. [tex]95.2 \, mol[/tex]
B. [tex]61.8 \, mol[/tex]
C. [tex]4.43 \, mol[/tex]
D. [tex]1.34 \, mol[/tex]

Sagot :

To determine how many moles of chlorine ([tex]\(\text{Cl}_2\)[/tex]) can be produced by decomposing 157 grams of sodium chloride ([tex]\(\text{NaCl}\)[/tex]), we will perform the following steps:

### Step 1: Calculate the number of moles of [tex]\(\text{NaCl}\)[/tex]

The mass of [tex]\(\text{NaCl}\)[/tex] given is 157 grams. The molar mass of [tex]\(\text{NaCl}\)[/tex] (sum of the atomic masses of sodium (Na) and chlorine (Cl)) is:
[tex]\[ \text{Molar mass of NaCl} = 22.99 \, \text{g/mol} + 35.45 \, \text{g/mol} = 58.44 \, \text{g/mol} \][/tex]

Using the molar mass, we can calculate the number of moles of [tex]\(\text{NaCl}\)[/tex]:
[tex]\[ \text{Moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} \][/tex]
[tex]\[ \text{Moles of NaCl} = \frac{157 \, \text{g}}{58.44 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of NaCl} \approx 2.686 \, \text{mol} \][/tex]

### Step 2: Determine the moles of [tex]\(\text{Cl}_2\)[/tex] produced from the moles of [tex]\(\text{NaCl}\)[/tex]

The balanced chemical equation is:
[tex]\[ 2 \, \text{NaCl} \rightarrow 2 \, \text{Na} + \text{Cl}_2 \][/tex]

From the balanced equation, we see that 2 moles of [tex]\(\text{NaCl}\)[/tex] produce 1 mole of [tex]\(\text{Cl}_2\)[/tex]. Therefore, the relationship between moles of [tex]\(\text{NaCl}\)[/tex] and moles of [tex]\(\text{Cl}_2\)[/tex] is:
[tex]\[ \text{Moles of Cl}_2 = \frac{\text{Moles of NaCl}}{2} \][/tex]

Using this relationship, we can find the moles of [tex]\(\text{Cl}_2\)[/tex] produced:
[tex]\[ \text{Moles of Cl}_2 = \frac{2.686 \, \text{mol}}{2} \][/tex]
[tex]\[ \text{Moles of Cl}_2 \approx 1.343 \, \text{mol} \][/tex]

Thus, approximately 1.34 moles of chlorine ([tex]\(\text{Cl}_2\)[/tex]) can be produced by decomposing 157 grams of sodium chloride ([tex]\(\text{NaCl}\)[/tex]).

So, the correct answer is:
D. [tex]\(1.34 \, \text{mol}\)[/tex]