calculate the mass of Pbl2 produced by reacting of 30.0g Kl with excess Pb( NO3)2
To calculate the mass of Pbl2 produced, we need to first set up a balanced chemical equation for the reaction between Kl (potassium iodide) and Pb(NO3)2 (lead(II) nitrate):
2 Kl + Pb(NO3)2 -> 2 KNO3 + PbI2
From the equation, we can see that for every 2 moles of Kl, we produce 1 mole of PbI2. To determine the mass of PbI2 produced, we'll follow these steps:
Step 1: Convert the given mass of Kl to moles.
To do this, we need to know the molar mass of Kl.
The molar mass of K (potassium) is 39.10 g/mol, and the molar mass of I (iodine) is 126.90 g/mol.
So, the molar mass of Kl is:
39.10 g/mol (K) + 126.90 g/mol (I) = 166.00 g/mol
To find the moles of Kl, we divide the given mass by the molar mass:
30.0 g Kl / 166.00 g/mol = 0.1807 mol Kl
Step 2: Determine the mole ratio between Kl and PbI2.
From the balanced chemical equation, we see that the mole ratio between Kl and PbI2 is 2:1.
Step 3: Calculate the moles of PbI2 produced.
Since the mole ratio between Kl and PbI2 is 2:1, the number of moles of PbI2 is equal to half the number of moles of Kl:
0.1807 mol Kl * (1 mol PbI2 / 2 mol Kl) = 0.0904 mol PbI2
Step 4: Convert the moles of PbI2 to grams.
To do this, we need to know the molar mass of PbI2.
The molar mass of Pb (lead) is 207.2 g/mol, and the molar mass of I (iodine) is 126.90 g/mol.
So, the molar mass of PbI2 is:
207.2 g/mol (Pb) + 2(126.90 g/mol (I)) = 461.0 g/mol
To find the mass of PbI2, we multiply the moles by the molar mass:
0.0904 mol PbI2 * 461.0 g/mol = 41.6 g PbI2
Therefore, the mass of PbI2 produced by reacting 30.0 g Kl with excess Pb(NO3)2 is 41.6 grams.
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3
mols KI = grams/molar mass = ?
mols PbI2 = 1/2 mols KI
mass PbI2 = mols PbI2 x molar mass PbI2