Pb(NO3)2 6.0mL 0.5 mol
NaI 12.0mL 0.5 mol
What is the mole ratio between Pb(NO3)2 and NaI?
You need to be more specific about your question.
I have the same lab. The ratio is 1:2.
Pb(NO3)2 being 1. NaI being the 2.
To determine the mole ratio between Pb(NO3)2 and NaI, we need to compare the number of moles of each compound.
Given:
Pb(NO3)2: Volume = 6.0 mL, Moles = 0.5 mol
NaI: Volume = 12.0 mL, Moles = 0.5 mol
First, let's convert the given volumes from milliliters (mL) to liters (L).
Volume of Pb(NO3)2 = 6.0 mL = 6.0/1000 L = 0.006 L
Volume of NaI = 12.0 mL = 12.0/1000 L = 0.012 L
Now, we can calculate the moles of each compound per liter.
Moles of Pb(NO3)2 per liter = (0.5 mol) / (0.006 L) = 83.33 mol/L
Moles of NaI per liter = (0.5 mol) / (0.012 L) = 41.67 mol/L
The mole ratio compares the number of moles of two substances in a chemical equation. From the calculations above, the mole ratio between Pb(NO3)2 and NaI is approximately 83.33:41.67, which can be simplified to 5:2.
Therefore, the mole ratio between Pb(NO3)2 and NaI is 5:2.