1. Consider the following reaction:
Pb(NO3)2+KI--> Pbl2 +KNO3
in theis experiment, 12.0g of Kl is reacted with 20.5g of Pb(NO3)2.
a) write a balanced equation for this reaction.
b)identify the limiting reaction.
c)calculate the mass of KNO3 produced from this reaction.
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
b. mols KI = 12.0/molar mass KI.
mols Pb(NO3)2 = 20.5/molar mass Pb(NO)3)2.
Convert mols KI to mols KNO3 using the coefficients in the balanced equation.
Do the same for Pb(NO3)2 to mols KNO3.
It is likely these two valus will not agree which means one of them is not right; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
c. Using the smaller value for mols KNO3, g = mols x molar mass. = grams KNO3.
dear sir, can you give example more specific...i'm struggle hard to understand it but still not get it..i'm strongly weak for this subject..plz
a) To balance the equation, we need to ensure that there are an equal number of atoms on both sides. First, let's write the chemical formulas for the compounds involved:
Lead(II) nitrate: Pb(NO3)2
Potassium iodide: KI
Lead(II) iodide: PbI2
Potassium nitrate: KNO3
Now, let's balance the equation:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
b) To identify the limiting reactant, we need to determine which reactant is present in the lowest amount and use stoichiometry to calculate how much product could be formed using each reactant. We will compare the amount of product formed from each reactant and choose the reactant that produces the least amount of product.
First, let's calculate the moles of each reactant:
Given:
Mass of KI = 12.0g
Mass of Pb(NO3)2 = 20.5g
To calculate the moles, we use the molar mass of each compound:
Molar mass of KI = 39.10 + 126.90 = 166.00 g/mol
Molar mass of Pb(NO3)2 = 207.20 + (3 * 16.00) = 331.20 g/mol
Moles of KI = (12.0g / 166.00 g/mol) = 0.0723 mol
Moles of Pb(NO3)2 = (20.5g / 331.20 g/mol) = 0.0619 mol
Now, let's determine the moles of product formed from each reactant:
From the balanced equation, we can see that 1 mole of Pb(NO3)2 produces 1 mole of PbI2 and 2 moles of KNO3.
Moles of PbI2 formed from Pb(NO3)2 = Moles of Pb(NO3)2 = 0.0619 mol
Moles of KNO3 formed from Pb(NO3)2 = 2 * Moles of Pb(NO3)2 = 2 * 0.0619 mol = 0.1238 mol
Moles of KNO3 formed from KI = Moles of KI / 2 = 0.0723 mol / 2 = 0.03615 mol
By comparing the moles of product formed, we see that the limiting reactant is Pb(NO3)2 because it produces the least amount of KNO3.
c) To calculate the mass of KNO3 produced, we will use the molar mass of KNO3:
Molar mass of KNO3 = 39.10 + 14.01 + (3 * 16.00) = 101.10 g/mol
Mass of KNO3 formed from Pb(NO3)2 = Moles of KNO3 * Molar mass of KNO3
= 0.1238 mol * 101.10 g/mol
= 12.52 g
Therefore, the mass of KNO3 produced from this reaction is 12.52 grams.