Answer:
1. 2.80g Pb +2
2. 3.34g PbF2
Step-by-step explanation:
The molar mass of lead (Pb) is approximately 207.2 g/mol.
The molar mass of fluorine (F) is approximately 19.0 g/mol.
To calculate the molar mass of PbF2, we add the atomic masses of lead and two fluorine atoms:
Molar mass of PbF2 = (207.2 g/mol) + 2 * (19.0 g/mol) = 245.2 g/mol
The mol ratio of PbF2 is:
for every 1 pb +2, there are 2 F -
1 Pb : 2 F
1. How many grams of lead(II) are present in 3.31 grams of lead(II) fluoride, PbF2?
To determine the amount of lead in 3.31 grams of PbF2, we need to calculate the mass ratio between lead and lead(II) fluoride.
(207.2 g Pb / 245.2 g PbF2) * 3.31 g PbF2 = 2.80 grams of lead (II).
2. How many grams of lead(II) fluoride can be made from 2.82 grams of lead(II)?
To determine the amount of lead(II) fluoride that can be made from 2.82 grams of lead, we use the mass ratio between lead and lead(II) fluoride.
(245.2 g PbF2 / 207.2 g Pb) * 2.82 g Pb = 3.34 grams of lead(II) fluoride.
In the Photo below, you can see how I used a Ratio and the train track method to get and defend my answers.
Hope this helps!