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 produced when chlorine reacts with 0.29 g of sodium iodine, we need to first write the balanced equation for the reaction.
The reaction between chlorine (Cl₂) and sodium iodide (NaI) is a single replacement reaction, which can be represented as:
2NaI + Cl₂ → 2NaCl + I₂
From the balanced equation, we can see that two moles of sodium iodide (NaI) react with one mole of chlorine (Cl₂) to produce two moles of sodium chloride (NaCl).
Now, we need to convert the given mass of sodium iodide to moles. To do this, we use the molar mass of sodium iodide (NaI), which can be calculated by adding the atomic masses of its constituent elements:
Na (sodium) = 22.99 g/mol
I (iodine) = 126.90 g/mol
Molar mass of NaI = (22.99 g/mol) + (126.90 g/mol) = 149.89 g/mol
To convert grams to moles, we divide the given mass by the molar mass:
Moles of NaI = 0.29 g / 149.89 g/mol
Now, we can use stoichiometry to determine the moles of sodium chloride (NaCl) produced. From the balanced equation, we know that for every 2 moles of sodium iodide reacting, we get 2 moles of sodium chloride:
Moles of NaCl = (Moles of NaI) x (2 moles of NaCl / 2 moles of NaI)
Finally, we can convert moles of sodium chloride to grams by multiplying by the molar mass of sodium chloride (NaCl), which is 58.44 g/mol:
Mass of NaCl = (Moles of NaCl) x (58.44 g/mol)
By following these calculations, you can determine the mass of sodium chloride produced when chlorine reacts with 0.29 g of sodium iodide in a single replacement reaction.