Question
If 40g of Sodium (Na) completely reacts with 60g of Chlorine (Cl2), what is the mass of the products? O a О b • с O d 40 g 60 g 100 g 160 g
Answers
Answered by
GPT-4o mini
To determine the mass of the products formed from the reaction between sodium (Na) and chlorine (Cl\(_2\)), we first need to recognize the reaction that occurs:
\[ 2 \, \text{Na} + \text{Cl}_2 \rightarrow 2 \, \text{NaCl} \]
In this reaction, 2 moles of sodium react with 1 mole of chlorine gas to produce 2 moles of sodium chloride (NaCl).
1. **Calculate moles of sodium and chlorine:**
- Molar mass of sodium (Na) is approximately 23 g/mol.
- Molar mass of chlorine (Cl\(_2\)) is approximately 70.9 g/mol (2 x 35.45 g/mol).
Moles of sodium:
\[
\text{Moles of Na} = \frac{40 \, \text{g}}{23 \, \text{g/mol}} \approx 1.74 \, \text{mol}
\]
Moles of chlorine:
\[
\text{Moles of Cl}_2 = \frac{60 \, \text{g}}{70.9 \, \text{g/mol}} \approx 0.85 \, \text{mol}
\]
2. **Determine the limiting reactant:**
According to the balanced equation, 2 moles of Na reacts with 1 mole of Cl\(_2\). So, we need to consider the ratio:
- From the available moles of Na: \( 1.74 \, \text{mol Na} \) would require \( \frac{1.74}{2} = 0.87 \, \text{mol Cl}_2 \).
- We only have 0.85 mol Cl\(_2\), which means Cl\(_2\) is the limiting reactant.
3. **Calculate the mass of the products:**
- Since Cl\(_2\) is the limiting reactant, we will use it to find the amount of NaCl produced. For every 1 mole of Cl\(_2\), 2 moles of NaCl are produced.
Moles of NaCl produced:
\[
\text{Moles of NaCl} = 2 \times 0.85 \, \text{mol Cl}_2 = 1.7 \, \text{mol NaCl}
\]
Calculating the mass of NaCl produced:
- Molar mass of NaCl is approximately 58.5 g/mol.
Mass of NaCl:
\[
\text{Mass of NaCl} = 1.7 \, \text{mol} \times 58.5 \, \text{g/mol} \approx 99.45 \, \text{g}
\]
4. **Total mass of reactants:**
\[
\text{Total mass of reactants} = 40 \, \text{g (Na)} + 60 \, \text{g (Cl}_2) = 100 \, \text{g}
\]
According to the law of conservation of mass, the mass of the products will equal the mass of the reactants. Thus, the mass of the products is:
**100 g**.
The correct answer is **100 g** (option c).
\[ 2 \, \text{Na} + \text{Cl}_2 \rightarrow 2 \, \text{NaCl} \]
In this reaction, 2 moles of sodium react with 1 mole of chlorine gas to produce 2 moles of sodium chloride (NaCl).
1. **Calculate moles of sodium and chlorine:**
- Molar mass of sodium (Na) is approximately 23 g/mol.
- Molar mass of chlorine (Cl\(_2\)) is approximately 70.9 g/mol (2 x 35.45 g/mol).
Moles of sodium:
\[
\text{Moles of Na} = \frac{40 \, \text{g}}{23 \, \text{g/mol}} \approx 1.74 \, \text{mol}
\]
Moles of chlorine:
\[
\text{Moles of Cl}_2 = \frac{60 \, \text{g}}{70.9 \, \text{g/mol}} \approx 0.85 \, \text{mol}
\]
2. **Determine the limiting reactant:**
According to the balanced equation, 2 moles of Na reacts with 1 mole of Cl\(_2\). So, we need to consider the ratio:
- From the available moles of Na: \( 1.74 \, \text{mol Na} \) would require \( \frac{1.74}{2} = 0.87 \, \text{mol Cl}_2 \).
- We only have 0.85 mol Cl\(_2\), which means Cl\(_2\) is the limiting reactant.
3. **Calculate the mass of the products:**
- Since Cl\(_2\) is the limiting reactant, we will use it to find the amount of NaCl produced. For every 1 mole of Cl\(_2\), 2 moles of NaCl are produced.
Moles of NaCl produced:
\[
\text{Moles of NaCl} = 2 \times 0.85 \, \text{mol Cl}_2 = 1.7 \, \text{mol NaCl}
\]
Calculating the mass of NaCl produced:
- Molar mass of NaCl is approximately 58.5 g/mol.
Mass of NaCl:
\[
\text{Mass of NaCl} = 1.7 \, \text{mol} \times 58.5 \, \text{g/mol} \approx 99.45 \, \text{g}
\]
4. **Total mass of reactants:**
\[
\text{Total mass of reactants} = 40 \, \text{g (Na)} + 60 \, \text{g (Cl}_2) = 100 \, \text{g}
\]
According to the law of conservation of mass, the mass of the products will equal the mass of the reactants. Thus, the mass of the products is:
**100 g**.
The correct answer is **100 g** (option c).