determine the mass of sodium chloride produced when chlorine reacts with 0.29 g of sodium iodine in a single replacement reaction
To determine the mass of sodium chloride (NaCl) produced when chlorine (Cl₂) reacts with 0.29 g of sodium iodide (NaI) in a single replacement reaction, we first need to write out a balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between chlorine and sodium iodide is:
2NaI + Cl₂ → 2NaCl + I₂
According to the balanced equation, 2 moles of sodium iodide react with 1 mole of chlorine to produce 2 moles of sodium chloride.
1 mole of sodium iodide (NaI) has a molar mass of 149.89 g/mol (sodium: 22.99 g/mol + iodine: 126.90 g/mol).
To find the number of moles of sodium iodide used, we can use the equation:
moles = mass / molar mass
moles of NaI = 0.29 g / 149.89 g/mol = 0.001934 mol
Since 2 moles of sodium iodide react with 1 mole of chlorine to produce 2 moles of sodium chloride, the number of moles of chlorine is also 0.001934 mol.
Now, using the balanced equation, we can determine the number of moles of sodium chloride produced:
moles of NaCl = 2 moles of Cl₂ (from the balanced equation) = 2 * 0.001934 mol = 0.003868 mol
Finally, to find the mass of sodium chloride produced, we can use the equation:
mass = moles * molar mass
mass of NaCl = 0.003868 mol * (sodium: 22.99 g/mol + chlorine: 35.45 g/mol) = 0.1987 g
Therefore, the mass of sodium chloride produced when chlorine reacts with 0.29 g of sodium iodide is approximately 0.1987 g.