Answer:
a) To find the limiting reactant, we need to calculate the number of moles of each reactant and compare their ratios to the coefficients in the balanced equation.
Moles of Pb(NO3)2 = 5
Moles of KI = 8
From the balanced equation, the ratio of Pb(NO3)2 to KI is 1:2. Therefore, Pb(NO3)2 is the limiting reactant since we have less than half the number of moles needed to react with all of the KI. KI is the excess reactant.
b) To find the grams of the excess reactant used, we need to convert the number of moles of KI to grams using its molar mass.
Molar mass of KI = 39.1 + 126.9 = 166 g/mol
Moles of KI = 8
Mass of KI = 8 mol x 166 g/mol = 1328 g
c) Since KI is the excess reactant, not all of it will be used up in the reaction. To find the amount remaining, we need to subtract the amount used from the initial amount.
Initial mass of KI = 8 mol x 166 g/mol = 1328 g
Mass of KI used = mass of Pbl₂ produced/2 (from the balanced equation)
Theoretical yield of Pbl₂ = 5 mol x 461 g/mol = 2305 g
Mass of KI used = 2305 g/2 = 1152.5 g
Mass of KI remaining = 1328 g - 1152.5 g = 175.5 g
d) The balanced equation tells us that 1 mole of Pb(NO3)2 produces 1 mole of Pbl₂. Therefore, the number of moles of Pbl₂ produced is equal to the number of moles of Pb(NO3)2 used.
Moles of Pb(NO3)2 = 5
Molar mass of Pbl₂ = 207.2 + 2(126.9) = 461 g/mol
Theoretical yield of Pbl₂ = 5 mol x 461 g/mol = 2305 g
e) The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Percent yield = (actual yield/theoretical yield) x 100
Percent yield = (1800 g/2305 g) x 100 = 78.05%
Step-by-step explanation:
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