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A chemist reacted 15.0 liters of [tex]$F_2$[/tex] gas with NaCl in the laboratory to form [tex]$Cl_2$[/tex] and NaF. Use the ideal gas law equation to determine the mass of NaCl that reacted with [tex][tex]$F_2$[/tex][/tex] at [tex]280 \, K[/tex] and 1.50 atm.

[tex]F_2 + 2 \, NaCl \rightarrow Cl_2 + 2 \, NaF[/tex]

Sagot :

GinnyAnsweridn
Certainly! Let's go through the process step-by-step.

### Step 1: Identify the Given Information
We are given:
- Volume of [tex]\( F_2 \)[/tex] gas ([tex]\( V \)[/tex]) = 15.0 liters
- Temperature ([tex]\( T \)[/tex]) = 280 K
- Pressure ([tex]\( P \)[/tex]) = 1.50 atm
- Ideal Gas Constant ([tex]\( R \)[/tex]) = 0.0821 L·atm/(mol·K)

### Step 2: Ideal Gas Law
We need to use the Ideal Gas Law, which is given by:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure
- [tex]\( V \)[/tex] is the volume
- [tex]\( n \)[/tex] is the number of moles of gas
- [tex]\( R \)[/tex] is the ideal gas constant
- [tex]\( T \)[/tex] is the temperature

### Step 3: Calculate the Number of Moles of [tex]\( F_2 \)[/tex] Gas
Rearranging the Ideal Gas Law to solve for [tex]\( n \)[/tex], we get:
[tex]\[ n = \frac{PV}{RT} \][/tex]

Plugging in the given values:
[tex]\[ n_{F_2} = \frac{(1.50 \, \text{atm}) \times (15.0 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K)}) \times (280 \, \text{K})} \][/tex]

Upon performing the calculation:
[tex]\[ n_{F_2} \approx 0.9788 \, \text{moles} \][/tex]

### Step 4: Stoichiometric Relationships
From the balanced chemical equation:
[tex]\[ F_2 + 2 NaCl \rightarrow Cl_2 + 2 NaF \][/tex]

This tells us that 1 mole of [tex]\( F_2 \)[/tex] reacts with 2 moles of [tex]\( NaCl \)[/tex]. So, we need to find how many moles of [tex]\( NaCl \)[/tex] react with [tex]\( n_{F_2} \)[/tex] moles of [tex]\( F_2 \)[/tex]:
[tex]\[ \text{Moles of } NaCl = 2 \times n_{F_2} \][/tex]

Substituting the moles of [tex]\( F_2 \)[/tex]:
[tex]\[ \text{Moles of } NaCl = 2 \times 0.9788 \approx 1.9575 \, \text{moles} \][/tex]

### Step 5: Calculate the Mass of [tex]\( NaCl \)[/tex]
Next, we need to convert moles of [tex]\( NaCl \)[/tex] to grams. The molar mass of [tex]\( NaCl \)[/tex] is:
[tex]\[ \text{Molar Mass of } NaCl = 22.99 \, \text{g/mol (Na)} + 35.45 \, \text{g/mol (Cl)} = 58.44 \, \text{g/mol} \][/tex]

So, the mass of [tex]\( NaCl \)[/tex] that reacted:
[tex]\[ \text{Mass of } NaCl = \text{Moles of } NaCl \times \text{Molar Mass of } NaCl \][/tex]
[tex]\[ \text{Mass of } NaCl \approx 1.9575 \, \text{moles} \times 58.44 \, \text{g/mol} \approx 114.4 \, \text{grams} \][/tex]

### Conclusion
Thus, the mass of [tex]\( NaCl \)[/tex] that reacted with [tex]\( F_2 \)[/tex] under the given conditions is approximately 114.4 grams.
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