Given this equation:
2KI + Pb(NO3)2 ---> PbI2 + 2KNO3
calculate mass of PbI2 produced by reacting of 30.0 g KI with excess Pb(NO3)2
Please help
30g KI = 0.18 moles
You will get 1/2 that many moles of PbI2
so how many grams is that?
To calculate the mass of PbI2 produced, we need to use stoichiometry and the given mass of KI.
1. Determine the molar mass of KI and PbI2:
- The molar mass of KI (potassium iodide) can be found by adding the atomic masses of potassium (K) and iodine (I), which are 39.10 g/mol and 126.90 g/mol, respectively.
Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
- The molar mass of PbI2 (lead iodide) can be calculated by adding the atomic mass of lead (Pb) and twice the atomic mass of iodine (I).
Molar mass of PbI2 = 207.20 g/mol + (2 * 126.90 g/mol) = 459.00 g/mol
2. Convert the given mass of KI to moles:
- Use the molar mass of KI (166.00 g/mol) to convert grams to moles:
Moles of KI = Mass of KI / Molar mass of KI
= 30.0 g / 166.00 g/mol
3. Use the stoichiometry of the balanced equation to determine the moles of PbI2 produced:
According to the balanced equation: 2 moles of KI react to form 1 mole of PbI2.
So, the moles of PbI2 produced = (Moles of KI) / 2
4. Calculate the mass of PbI2 produced:
- Use the molar mass of PbI2 (459.00 g/mol) to convert moles to grams:
Mass of PbI2 = Moles of PbI2 * Molar mass of PbI2
Let's perform the calculations:
Molar mass of KI = 166.00 g/mol
Molar mass of PbI2 = 459.00 g/mol
Moles of KI = 30.0 g / 166.00 g/mol
Moles of PbI2 = (Moles of KI) / 2
Mass of PbI2 = Moles of PbI2 * Molar mass of PbI2
Please substitute the values and calculate the final answer.
To calculate the mass of PbI2 produced, you need to use the equation coefficients to determine the mole ratio between KI and PbI2. Then, you can convert the number of moles of KI to moles of PbI2 using the mole ratio. Finally, you can convert the moles of PbI2 to grams using the molar mass of PbI2.
Here's the step-by-step solution:
Step 1: Determine the molar mass of KI and PbI2.
- The molar mass of KI is the sum of the atomic masses of potassium (K) and iodine (I) which is 39.10 g/mol (K) + 126.90 g/mol (I) = 166.00 g/mol.
- The molar mass of PbI2 is the sum of the atomic mass of lead (Pb) and twice the atomic mass of iodine (I) which is 207.20 g/mol (Pb) + 2 * 126.90 g/mol (I) = 461.00 g/mol.
Step 2: Calculate the moles of KI.
- Given the mass of KI is 30.0 g.
- Convert grams to moles using the molar mass of KI: 30.0 g KI * (1 mol KI / 166.00 g KI) = 0.181 moles of KI.
Step 3: Determine the mole ratio of KI to PbI2 from the balanced equation.
- According to the balanced equation: 2 moles of KI react to produce 1 mole of PbI2.
- Therefore, the mole ratio of KI to PbI2 is 2:1.
Step 4: Calculate the moles of PbI2 produced.
- Use the mole ratio from step 3 to convert the moles of KI to moles of PbI2: 0.181 moles of KI * (1 mole PbI2 / 2 moles KI) = 0.0905 moles of PbI2.
Step 5: Convert moles of PbI2 to grams.
- Multiply the moles of PbI2 by the molar mass of PbI2: 0.0905 moles PbI2 * 461.00 g/mol PbI2 = 41.7515 grams of PbI2.
Therefore, the mass of PbI2 produced by reacting 30.0 g of KI with excess Pb(NO3)2 is approximately 41.75 grams.