how many grams of potassium iodide can be produced by the reaction of 150 grams of iodide and 45.6 grams of potassium?
This is a limiting reagent (LR) problem.
I know that because amounts are given for BOTH reactants.
2K | i2 ==> 2kI
mols K = grams/molar mass
mols I2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols K to mols KI.
Do the same for mols I2 to mols KI. It is likely that the values will not agree; the correct value in LR problem is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value for mols KI, convert to g. g = mols x molar mass.
To determine the grams of potassium iodide produced in the reaction, we need to understand the balanced chemical equation for the reaction between iodide and potassium.
The balanced equation for the reaction is:
2 K + 2 I → 2 KI
From the balanced equation, we can see that for every 2 moles of potassium (K), we produce 2 moles of potassium iodide (KI). Therefore, the stoichiometric ratio between potassium and potassium iodide is 1:1.
To calculate the moles of potassium iodide produced, we need to convert the masses of iodide and potassium to moles, using their molar masses.
The molar mass of iodide (I) is 127 g/mol, and the molar mass of potassium (K) is 39.1 g/mol.
Given:
Mass of iodide (I) = 150 grams
Mass of potassium (K) = 45.6 grams
Calculations:
Moles of iodide (I) = mass / molar mass = 150 g / 127 g/mol = 1.18 mol
Moles of potassium (K) = mass / molar mass = 45.6 g / 39.1 g/mol = 1.17 mol
Since the stoichiometric ratio between potassium and potassium iodide is 1:1, the moles of potassium iodide produced will be equal to the moles of iodide or potassium.
Therefore, the moles of potassium iodide produced = 1.18 mol (or 1.17 mol)
Finally, we can calculate the mass of potassium iodide by multiplying the moles by the molar mass of potassium iodide.
The molar mass of potassium iodide (KI) is 166 g/mol.
Mass of potassium iodide = moles * molar mass = 1.18 mol * 166 g/mol = 195.48 grams (or 1.17 mol * 166 g/mol = 193.62 grams)
Therefore, approximately 195.48 grams (or 193.62 grams) of potassium iodide can be produced.