Final answer:
The mass of sodium that reacted with 0.142 kg of chlorine to form 234 g of sodium chloride is 92.0 g. This was determined by using the stoichiometry of the balanced chemical reaction and converting grams to moles and back to grams.
Step-by-step explanation:
To find the mass of sodium that reacted, we need to use the stoichiometry of the balanced chemical equation for the reaction between sodium and chlorine to form sodium chloride (NaCl). The balanced equation is:
2 Na(s) + Cl₂(g) → 2 NaCl(s)
From the equation, we see that two moles of sodium react with one mole of chlorine to produce two moles of sodium chloride. We can use the provided masses to calculate the moles of chlorine and sodium chloride, and then use stoichiometry to find the mass of sodium that reacted.
First, we will convert the mass of chlorine into moles using its molar mass:
-
- 0.142 kg of Cl₂ = 142 g Cl₂
-
- The molar mass of Cl₂ (Cl-Cl) is approximately 70.90 g/mol.
-
- Moles of Cl₂ = 142 g / 70.90 g/mol ≈ 2.001 moles of Cl₂
Next, we'll use the stoichiometry of the reaction to find the moles of sodium. According to the balanced equation, the number of moles of sodium is twice that of chlorine since the stoichiometric coefficient of sodium is 2.
-
- Moles of Na = 2 × moles of Cl₂ = 2 × 2.001 moles ≈ 4.002 moles of Na
Then we'll convert moles of sodium to grams using the atomic mass of sodium:
-
- The atomic mass of Na is approximately 22.99 g/mol.
-
- Mass of Na = 4.002 moles × 22.99 g/mol ≈ 92.02 g
Therefore, the mass of sodium that reacted is 92.0 g, to three significant figures.