In the Reaction: Pb(NO3)2 + 2 Kl —> Pbl2 + 2 KNO3 How many grams of lead (II) iodide, Pbl2, is produced from 82.7 grams of potassium iodide, KI?

Round to two decimal places.

114.8g PbI2 are produced

Explanation:

To solve this question we must convert the mass of KI to moles using its molar mass (KI = 166.0g/mol). As 2 moles of KI produce 1 mole of PbI2 we can find its moles and its mass (Molar mass PbI2: 461.01g/mol) as follows:

Moles KI:

82.7g KI * (1mol / 166.0g) = 0.498 moles KI

Moles PbI2:

0.498 moles KI * (1mol PbI2 / 2mol KI) = 0.249 moles PbI2

Mass PbI2:

0.249moles * (461.01g/mol) =

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )