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).
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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: \[ \text{Moles of Cl}_2 = \frac{60 , \text{g}}{70.9 , \text{g/mol}} \approx 0.85 , \text{mol} \]
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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.
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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: \[ \text{Mass of NaCl} = 1.7 , \text{mol} \times 58.5 , \text{g/mol} \approx 99.45 , \text{g} \]
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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).