To solve this problem, we first need to determine the limiting reactant.
1. Calculate the molar mass of sodium iodide (NaI):
Na: 22.99 g/mol
I: 126.9 g/mol
Molar mass of NaI = 22.99 + 126.9 = 149.89 g/mol
2. Calculate the number of moles of sodium iodide:
moles = mass / molar mass
moles = 0.29 g / 149.89 g/mol
moles = 0.001934 moles
3. Use the stoichiometry of the balanced equation to determine the theoretical yield of sodium chloride:
From the balanced equation, 1 mole of NaI reacts with 1 mole of Cl2 to produce 2 moles of NaCl.
Therefore, the mole ratio of NaI to NaCl is 1:2.
moles of NaCl = 0.001934 moles x (2 moles NaCl / 1 mole NaI)
moles of NaCl = 0.003868 moles
4. Calculate the mass of sodium chloride produced:
mass = moles x molar mass
molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol
mass = 0.003868 moles x 58.44 g/mol
mass = 0.226 grams
Therefore, when 0.29 grams of sodium iodide reacts with chlorine gas, 0.226 grams of sodium chloride will be produced.
What mass of sodium chloride is produced when chlorine gas reacts with 0.29 grams of sodium iodide? The equation is NaI+Cl2=NaCl+I2
1 answer