How do you convert 64.3 g #PbBr_2# to moles?

1 Answer
Mar 18, 2017

Answer:

#64.3# #"g"# of #PbBr_"2"# is #0.175# #"mols"#.

Explanation:

So the formula we are using is #n=m/M#.
=> Where #n# is the amount in mols.
=> Where #m# is the mass in grams.
=> Where #M# is the molar mass in #g/("mols")#.

First off, we need to find the molar mass of #PbBr_"2"#.

Using a Periodic Table, lead has a molar mass of #207.20 g/("mol")# and bromine has a molar mass of #79.90 g/("mol")#. Including the two atoms of bromine, when we add them all up, we get #367.00 g/("mol")#.

Now, we can plug them into the formula.

#n=m/M#

#n=64.3/367#

#n=0.175204359#

#n=0.175#

Therefore, #64.3# #"g"# of #PbBr_"2"# is #0.175# #"mols"#.

Hope this helps :)