Question
f 40g of Sodium (Na) completely reacts with 60g of Chlorine (Cl2), what is the mass of the products?
a
40 g
b
60 g
c
100 g
d
160 g
a
40 g
b
60 g
c
100 g
d
160 g
Answers
GPT-4o mini
To determine the mass of the products produced from the reaction of sodium (Na) and chlorine (Cl₂), we will first look at the balanced chemical reaction between sodium and chlorine:
\[ 2 \, \text{Na} + \text{Cl}_2 \rightarrow 2 \, \text{NaCl} \]
From the equation:
- 2 moles of sodium (Na) react with 1 mole of chlorine gas (Cl₂) to produce 2 moles of sodium chloride (NaCl).
**Step 1: Calculate moles of reactants.**
- Molar mass of Sodium (Na) = 23 g/mol
- For 40 g of Na:
\[
\text{Moles of Na} = \frac{40 \, \text{g}}{23 \, \text{g/mol}} \approx 1.74 \, \text{mol}
\]
- Molar mass of Chlorine gas (Cl₂) = 70.9 g/mol (approximately 35.45 g/mol for one Cl atom)
- For 60 g of Cl₂:
\[
\text{Moles of Cl}_2 = \frac{60 \, \text{g}}{70.9 \, \text{g/mol}} \approx 0.845 \, \text{mol}
\]
**Step 2: Determine the limiting reactant.**
According to the balanced equation, 2 moles of Na react with 1 mole of Cl₂. Thus, to react with the available sodium (1.74 moles), we should have:
\[
\text{Required moles of Cl}_2 = \frac{1.74 \, \text{mol Na}}{2} = 0.87 \, \text{mol Cl}_2
\]
Since we only have 0.845 moles of Cl₂, chlorine is the limiting reactant.
**Step 3: Calculate the mass of products formed.**
Since chlorine is the limiting reactant, we will use the moles of Cl₂ to find out how much Na and Cl will produce NaCl.
From the balanced equation, 1 mole of Cl₂ produces 2 moles of NaCl.
So, 0.845 mol of Cl₂ will produce:
\[
\text{Moles of NaCl} = 0.845 \, \text{mol Cl}_2 \times 2 \approx 1.69 \, \text{mol NaCl}
\]
**Step 4: Calculate the mass of NaCl produced.**
Molar mass of Sodium Chloride (NaCl) = 23 g/mol (Na) + 35.45 g/mol (Cl) = 58.45 g/mol.
Now, calculate the total mass of NaCl produced:
\[
\text{Mass of NaCl} = 1.69 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 98.77 \, \text{g}
\]
**Step 5: Calculate the total mass of the products.**
According to the law of conservation of mass, the mass of the products equals the mass of the reactants:
\[
\text{Total mass of reactants} = 40 \, \text{g (Na)} + 60 \, \text{g (Cl}_2\text{)} = 100 \, \text{g}
\]
Thus, the mass of the products is also 100 g.
So, the answer to the question is:
**c) 100 g**
\[ 2 \, \text{Na} + \text{Cl}_2 \rightarrow 2 \, \text{NaCl} \]
From the equation:
- 2 moles of sodium (Na) react with 1 mole of chlorine gas (Cl₂) to produce 2 moles of sodium chloride (NaCl).
**Step 1: Calculate moles of reactants.**
- Molar mass of Sodium (Na) = 23 g/mol
- For 40 g of Na:
\[
\text{Moles of Na} = \frac{40 \, \text{g}}{23 \, \text{g/mol}} \approx 1.74 \, \text{mol}
\]
- Molar mass of Chlorine gas (Cl₂) = 70.9 g/mol (approximately 35.45 g/mol for one Cl atom)
- For 60 g of Cl₂:
\[
\text{Moles of Cl}_2 = \frac{60 \, \text{g}}{70.9 \, \text{g/mol}} \approx 0.845 \, \text{mol}
\]
**Step 2: Determine the limiting reactant.**
According to the balanced equation, 2 moles of Na react with 1 mole of Cl₂. Thus, to react with the available sodium (1.74 moles), we should have:
\[
\text{Required moles of Cl}_2 = \frac{1.74 \, \text{mol Na}}{2} = 0.87 \, \text{mol Cl}_2
\]
Since we only have 0.845 moles of Cl₂, chlorine is the limiting reactant.
**Step 3: Calculate the mass of products formed.**
Since chlorine is the limiting reactant, we will use the moles of Cl₂ to find out how much Na and Cl will produce NaCl.
From the balanced equation, 1 mole of Cl₂ produces 2 moles of NaCl.
So, 0.845 mol of Cl₂ will produce:
\[
\text{Moles of NaCl} = 0.845 \, \text{mol Cl}_2 \times 2 \approx 1.69 \, \text{mol NaCl}
\]
**Step 4: Calculate the mass of NaCl produced.**
Molar mass of Sodium Chloride (NaCl) = 23 g/mol (Na) + 35.45 g/mol (Cl) = 58.45 g/mol.
Now, calculate the total mass of NaCl produced:
\[
\text{Mass of NaCl} = 1.69 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 98.77 \, \text{g}
\]
**Step 5: Calculate the total mass of the products.**
According to the law of conservation of mass, the mass of the products equals the mass of the reactants:
\[
\text{Total mass of reactants} = 40 \, \text{g (Na)} + 60 \, \text{g (Cl}_2\text{)} = 100 \, \text{g}
\]
Thus, the mass of the products is also 100 g.
So, the answer to the question is:
**c) 100 g**