Given the following equation: 2 Naclo —> 2 NaCl + 3 Oz. How many grams of NaCl are produced when 80.0 grams of O, are produced? (medium) (Molar mass of oxygen: 16 g/mol) (Molar mass of sodium chloride: 58.45 g/mol) (medium)
The correct answer and explanation is :
The balanced chemical equation provided is:
[ 2 \, \text{NaClO}_3 \rightarrow 2 \, \text{NaCl} + 3 \, \text{O}_2 ]
This equation indicates that 2 moles of sodium chlorate (NaClO₃) decompose to produce 2 moles of sodium chloride (NaCl) and 3 moles of oxygen gas (O₂).
Given Data:
- Molar mass of oxygen (O₂): 16 g/mol.
- Molar mass of sodium chloride (NaCl): 58.45 g/mol.
- Mass of oxygen produced: 80.0 grams.
Objective: Determine the mass of NaCl produced when 80.0 grams of O₂ are produced.
Solution:
- Calculate moles of O₂ produced: Given that the molar mass of O₂ is 16 g/mol, the number of moles of O₂ produced can be calculated as: [ \text{moles of O}_2 = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} = \frac{80.0 \, \text{g}}{16 \, \text{g/mol}} = 5.0 \, \text{mol} ]
- Use the stoichiometric relationship to find moles of NaCl produced: From the balanced equation, the mole ratio of NaCl to O₂ is 2:3. This means that for every 3 moles of O₂ produced, 2 moles of NaCl are produced. Therefore, the moles of NaCl produced are: [ \text{moles of NaCl} = \left( \frac{2 \, \text{mol NaCl}}{3 \, \text{mol O}_2} \right) \times 5.0 \, \text{mol O}_2 = \frac{10.0}{3} \, \text{mol NaCl} \approx 3.33 \, \text{mol NaCl} ]
- Calculate the mass of NaCl produced: Using the molar mass of NaCl (58.45 g/mol), the mass of NaCl produced is: [ \text{mass of NaCl} = \text{moles of NaCl} \times \text{molar mass of NaCl} = 3.33 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 194.8 \, \text{g} ]
Conclusion:
Approximately 194.8 grams of sodium chloride (NaCl) are produced when 80.0 grams of oxygen gas (O₂) are produced, based on the given chemical equation and molar masses.
Note: The molar mass of oxygen (O₂) is actually 32 g/mol (16 g/mol per oxygen atom). Therefore, the correct calculation for moles of O₂ is:
[ \text{moles of O}_2 = \frac{80.0 \, \text{g}}{32 \, \text{g/mol}} = 2.5 \, \text{mol} ]
Using this value, the moles of NaCl produced would be:
[ \text{moles of NaCl} = \left( \frac{2 \, \text{mol NaCl}}{3 \, \text{mol O}_2} \right) \times 2.5 \, \text{mol O}_2 = \frac{5.0}{3} \, \text{mol NaCl} \approx 1.67 \, \text{mol NaCl} ]
And the mass of NaCl produced:
[ \text{mass of NaCl} = 1.67 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 97.5 \, \text{g} ]
Therefore, approximately 97.5 grams of sodium chloride (NaCl) are produced when 80.0 grams of oxygen gas (O₂) are produced.
Explanation:
This calculation demonstrates the application of stoichiometric principles, where the relationships between reactants and products in a chemical reaction are used to determine the quantities of substances involved. By understanding the mole ratios from the balanced equation and knowing the molar masses of the substances, one can calculate the amount of product formed from a given amount of reactant.