Potassium iodide is prepared by the reaction of hydroiodic acid and potassium hydrogen carbonate. water and carbon dioxide are also produced. in the reaction of 398 g HI and 318 g KHCO3, how many grams of KI are produced?
1. This is a limiting reagent problem.
2 Write the equation and balance it.
3a. Convert g HI and g KHCO3 to moles. moles = grams/molar mass.
3b. Using the coefficients in the balanced equation, convert moles HI and moles KHCO3 to moles KI.
3c. You will obtain two answers for moles KI. Obviously both can't be correct. The SMALLER number of moles is the correct answer and it will identify the limiting reagent.
4. Using the smaller number of moles KI from step 3c, convert moles to grams. grams = moles x molar mass.
Post your work if you get stuck.
To find out how many grams of KI are produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed.
Let's calculate the number of moles for each reactant:
1. Hydroiodic acid (HI):
Number of moles = mass / molar mass
Number of moles of HI = 398 g / (1 g/mol + 127 g/mol) ≈ 2.95 mol
2. Potassium hydrogen carbonate (KHCO3):
Number of moles = mass / molar mass
Number of moles of KHCO3 = 318 g / (39 g/mol + 12 g/mol + 48 g/mol + 16 g/mol + 3 x 16 g/mol) ≈ 2.38 mol
Now, we need to determine the limiting reactant.
The balanced chemical equation for the reaction is:
2 HI + KHCO3 -> KI + CO2 + H2O
From the equation, we can see that 2 moles of HI react with 1 mole of KHCO3 to produce 1 mole of KI.
By comparing the moles of HI and KHCO3, we see that the KHCO3 has less moles (2.38 mol) than HI (2.95 mol).
Therefore, KHCO3 is the limiting reactant.
To find the mass of KI produced, we use the mole ratio from the balanced equation:
1 mole of KHCO3 produces 1 mole of KI
Number of moles of KI = Number of moles of KHCO3 = 2.38 mol
Finally, we can calculate the mass of KI produced:
Mass of KI = number of moles × molar mass
Mass of KI = 2.38 mol × (39 g/mol + 127 g/mol)
Mass of KI ≈ 420.2 g
Therefore, approximately 420.2 grams of KI are produced in this reaction.
To determine the number of grams of potassium iodide (KI) produced, we need to use stoichiometry. Stoichiometry is a mathematical relationship between the quantities of reactants and products in a chemical reaction. It is based on the balanced chemical equation for the reaction.
Let's start by writing the balanced chemical equation for the reaction:
2 HI + K2CO3 → 2 KI + H2O + CO2
From the balanced equation, we can see that:
2 moles of HI react with 1 mole of K2CO3 to produce 2 moles of KI.
Now, let's calculate the number of moles of HI and KHCO3 given their respective masses:
Molar mass of HI = 1.0079 g/mol + 126.9045 g/mol = 127.9124 g/mol
Molar mass of KHCO3 = 39.0983 g/mol + 1.0079 g/mol + 12.0107 g/mol + 3(15.999 g/mol) = 100.1155 g/mol
Number of moles of HI = 398 g / 127.9124 g/mol = 3.1127 mol
Number of moles of KHCO3 = 318 g / 100.1155 g/mol = 3.1785 mol
Now, let's determine the limiting reagent, which is the reactant that is completely consumed in the reaction:
From the balanced equation, we can see that 2 moles of HI react with 1 mole of K2CO3. Therefore, the mole ratio between HI and KHCO3 is 2:1.
Since the mole ratio is 2:1 and we have slightly more moles of KHCO3 (3.1785 mol) than HI (3.1127 mol), KHCO3 is in excess.
Now, let's calculate the number of moles of KI produced:
Moles of KI = 2 moles of KI (per 2 moles of HI) * 3.1127 mol of HI = 3.1127 mol
Finally, to calculate the mass of KI produced, we need to multiply the number of moles by the molar mass of KI:
Mass of KI = 3.1127 mol * 166.0028 g/mol = 516.2 g
Therefore, approximately 516.2 grams of KI are produced.