How many grams of lead(IV) oxide would be found in 1.23 moles of lead(IV) oxide?

Question

Asap. Need help

Answer 1

Hi Marlene,

 

1) Find the formula from the name:

 

Pb4+ and O2- (swap and drop charges, then reduce) = PbO2

 

2) Calculate the molar mass from the periodic table

 

Pb = 207.2 g/mol

 

2 x O = 2 x 16 g/mol

 

Total = 239.2 g/mol

 

3) Multiply moles by molar mass to cancel 'mol' unit

 

1.23 mol x 239.2 g/mol = 294.216 g

 

4) Round answer to correct sig figs

 

294 g

 

Hope this helps!

Answer 2

Hello Roxana

 

 

Lead(IV) oxide is PbO₂

First we need the molecular weight of the atoms involved

Pb is 207.2

O is 15.9994

We have 2 oxygens for every lead so the mass of the molecule is

207.2 + 2(15.9994) = 239.1988

 

There are 239.1988 grams per mole of PbO₂

 

We have 1.23 moles

 

1.23 mol • 239.1988 g/mol = 294.215 grams