[demvp_demar]

We did a lab yesterday and here are the results:


Pb(NO₃)₂+ 2KI   → 2KNO₃ + PbI₂

We had 0.95g of Lead (II) Nitrate and 1.10g of Potassium Iodide.

I've converted Pb(NO₃)₂ to Moles which gave me 0.00286 moles.

1.10g of KI₂ are 0.00662 moles.

How do I find the limiting reagent and what are the steps specifically.

[AWK]

  Pb(NO₃)₂   +     2KI         → 2KNO₃ + PbI₂
0.00286 mol  0.00662 mol
What you can see?

[demvp_demar]

Yea I got that part. But do I have to divide by the coefficients after? Or can I always just look at the moles and decide?

[AWK]

0.00286 mol  0.00662 mol - starting reagents
         ?                ?           - after reaction - which is null, and which in excess

[chenbeier]

Or in other words.

In the Equation you have a ratio 1 to 2.

Combine and compare it with the given values.

[Babcock_Hall]

It is the possible moles of product that one can make that is the key.  In order to know this, you have to use the stoichiometric coefficients and obey the normal rules of units cancellation.