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
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.