How many moles of lead iodide are produced from 300. g of potassium iodide? Usie the reaction below.
PB(NO3)+ KI = Ni9OH)2 + NaNO3
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Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
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To determine the number of moles of lead iodide produced, we need to use the given mass of potassium iodide (KI) and the balanced chemical equation provided.
First, we need to write the balanced chemical equation correctly:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
The balanced equation shows that 1 mole of lead(II) nitrate (Pb(NO3)2) reacts with 2 moles of potassium iodide (KI) to produce 1 mole of lead iodide (PbI2) and 2 moles of potassium nitrate (KNO3).
Now let's calculate the moles of potassium iodide (KI) using its molar mass.
Molar mass of KI = atomic mass of K + atomic mass of I
= (39.10 g/mole) + (126.90 g/mole)
= 166.00 g/mole
Next, we can calculate the number of moles of KI:
Moles of KI = mass of KI / molar mass of KI
= 300 g / 166.00 g/mole
≈ 1.807 moles of KI
According to the balanced equation, the mole ratio between KI and PbI2 is 2:1. Therefore, the number of moles of lead iodide (PbI2) produced will be half the number of moles of KI:
Moles of PbI2 = 1.807 moles of KI / 2
≈ 0.904 moles of PbI2
Therefore, approximately 0.904 moles of lead iodide (PbI2) will be produced from 300 g of potassium iodide (KI).
To determine the number of moles of lead iodide produced from 300. g of potassium iodide, we need to use the balanced chemical equation provided:
Pb(NO3)2 + 2KI -> PbI2 + 2KNO3
First, we need to calculate the molar mass of potassium iodide (KI) and lead iodide (PbI2):
- The molar mass of potassium (K) is approximately 39.10 g/mol.
- The molar mass of iodine (I) is approximately 126.90 g/mol.
Adding them together, we find that the molar mass of KI is approximately 39.10 + (126.90) = 166.00 g/mol.
- The molar mass of lead (Pb) is approximately 207.20 g/mol.
- The molar mass of iodine (I) is approximately 126.90 g/mol.
Adding them together, we find that the molar mass of PbI2 is approximately 207.20 + (2 x 126.90) = 461.00 g/mol.
Now, we can calculate the number of moles of potassium iodide (KI) using its molar mass:
Number of moles = mass / molar mass
Number of moles of KI = 300. g / 166.00 g/mol
Next, we can use the stoichiometric coefficients from the balanced equation to determine the mole ratio between KI and PbI2. From the equation, we can see that 2 moles of KI react to produce 1 mole of PbI2.
Using the mole ratio, we can calculate the number of moles of PbI2 produced:
Number of moles of PbI2 = Number of moles of KI x (1 mole PbI2 / 2 moles KI)
Finally, substitute the value from the previous step into the equation to find the number of moles:
Number of moles of PbI2 = (300. g / 166.00 g/mol) x (1 mole PbI2 / 2 moles KI) = (1.807 mole KI) x (1 mole PbI2 / 2 moles KI)
Simplifying, we find:
Number of moles of PbI2 = 1.807 moles PbI2
Therefore, approximately 1.807 moles of lead iodide are produced from 300. g of potassium iodide.