anytime you get one of these how much of this or that reacts or is produced or even yield or % yield problems, think of these steps...
1) write a balanced equation
2) convert everything given to MOLES
3) determine limiting reagent
4) convert moles limiting reagent to moles other species
5) convert moles back to mass. this is THEORETICAL mass aka theoretical yield
6) % yield = actual recovered mass / theoretical mass x 100%
ok? can you apply those steps here? here are some hints...
*** 1 ***
3 Cu(OH)2 + 2 H3AsO4 --> Cu3(AsO4)2 + 6 H2O.... is already balanced
*** 2 ***
22.74 g Cu(OH)2 x ( 1 mole / ___ g Cu(OH)2) = ??? moles Cu(OH)2
19.42 g H3AsO4 x ( 1 mole / ___ H3AsO4) = ??? moles H3AsO4
plug in the molecular weights of Cu(OH)2 and H3AsO4...
*** 3 ****
from the balanced equation above... 3 moles Cu(OH)2 reacts with 2 moles H3AsO4.. right?
??? moles Cu(OH)2 x (2 moles H3AsO4 / 3 moles Cu(OH)2) = ??? moles H3SO4
that is the theoretical amount of H3AsO4 required to react with all of the Cu(OH)2 you started with. if the amount of H3AsO4 you calculated in step 2 is more than than, you have excess H3AsO4 and Cu(OH)2 is limiting the reaction. if less, H3AsO4 is limiting.
*** 4 ***
if Cu(OH)2 is limiting...
??? moles Cu(OH)2 x (2 moles H3AsO4 / 3 moles Cu(OH)2) = ??? moles H3AsO4)
if H3AsO4 is limiting...
??? moles H3AsO4 x (3 moles Cu(OH)2 / 2 moles H3AsO4) = ??? moles Cu(OH)2
and now you have the moles of the XS reagent needed to completely react with all the limiting reagent
find the XS moles...
*** 5 ***
conver XS moles to mass...right?
