Chemical Reaction Calculation: Reflecting on Mass Conservation
What can we learn from a chemical reaction involving the elements sodium and chlorine?
How can we determine the grams of chlorine gas that reacted when 15.6 g of sodium produced 39.7 g of sodium chloride?
Answer:
15.6 g of sodium and 24.1 g of chlorine react to produce 39.7 g of sodium chloride.
Explanation:
Given data:
Mass of sodium = 15.6 g
Mass of sodium chloride = 39.7 g
Mass of chlorine react = ?
Mass of sodium react = ?
Solution:
Balance chemical equation:
2Na + Cl₂ → 2NaCl
Now we will calculate the moles of sodium reacted.
Number of moles = mass /molar mass
Number of moles = 15.6 g / 23 g/mol
Number of moles = 0.68 mol
Now we will calculate the moles of sodium chloride produced,
Number of moles of NaCl = 39.7 / 58.5 g/mol
Number of moles of NaCl = 0.68 mol
Since 0.68 moles of sodium produced 0.68 moles of sodium chloride or we can say 2 moles of sodium produced 2 moles of sodium chloride.
Now we calculate the mass of chlorine reacted,
From balance chemical equation we compare the moles of chlorine and sodium
Na : Cl
2 : 1
0.68 : 1/2× 0.68 = 0.34 mol
Mass of chlorine reacted = moles × molar mass
Mass of chlorine reacted = 0.34 mol × 71 g/mol = 24.1 g
According to the law of conservation of mass,
2Na + Cl₂ → 2NaCl
15.6 g + 24.1 g = 39.7 g
15.6 g of sodium and 24.1 g of chlorine react to produce 39.7 g of sodium chloride.
Reflecting on a chemical reaction involving sodium and chlorine allows us to explore the concept of stoichiometry, which involves the quantitative relationships between reactants and products in a chemical reaction. In this case, the reaction between sodium and chlorine resulted in the formation of sodium chloride, known as table salt.
By applying stoichiometry principles and balancing the chemical equation for the reaction (2Na + Cl₂ → 2NaCl), we can determine how many grams of chlorine gas reacted when 15.6 g of sodium produced 39.7 g of sodium chloride. The calculation involves converting the given masses of sodium and sodium chloride into moles, and then using the stoichiometric ratios from the balanced equation to find the mass of chlorine gas that reacted.
Through this reflective analysis of the chemical reaction, we gain a deeper understanding of how mass is conserved in reactions, as demonstrated by the calculation that shows how 15.6 g of sodium and 24.1 g of chlorine combine to form 39.7 g of sodium chloride. This exploration highlights the importance of precision in measuring reactants and products, as well as the fundamental principles of chemical reactions and stoichiometry.