Stoichiometry

Lead nitrate reacts with potassium iodide to form a precipitate known as lead iodide according to the equation below.

Pb(NO3)2 + 2KI => PbI2 + 2KNO3

Apparatus:
10ml pipette
0.1M solutions of lead nitrate and potassium iodide
Funnel
Filling bulb
Distilled water
2 100ml beakers
Filter paper
Scales
Method
Weigh a piece of filter paper.
Pipette 10 mls of lead nitrate solution into a beaker.
Pipette 10 mls of potassium iodide solution into the same beaker and notice the formation of the yellow solid (lead iodide).
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Filter the mixture as shown on the right.

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Using distilled water rinse any remaining solid into the filter paper.

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Allow the filter paper and funnel to dry overnight.

Record the results in a table

 Item Mass (grams) Filter paper Filter paper and lead iodide Mass of lead iodide

Questions

What is the mass of lead iodide formed?
Calculate the number of moles of lead iodide formed.
Consider the balanced equation for this reaction
Pb(NO3)2 + 2KI => PbI2 + 2KNO3
For every mole of lead nitrate reacted how many mole of potassium iodide formed?
How many mole of potassium iodide reacted to form the mass of lead iodide ?
What mass of potassium iodide reacted?

Consider the experiment mentioned below.

Jonathon conducted this experiment. He used a scale that gives readings to two decimal places. He weighed a single piece of filter paper as shown on the right.

He then weighed the same filter paper with the lead iodide after it had been left to dry overnight.
What is the mass of lead iodide formed?
How many mole of lead iodide is this?

How many mole of lead nitrate reacted?
What mass of lead nitrate reacted?

 Item Mass (grams) Filter paper Filter paper and lead iodide Mass of lead iodide