How would we write the equation between #Na_3PO_4# and #Pb(NO_3)_2#, given #17.3g# of the former, and #12.4g# of the latter?

Question

1811 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-05T02:23:29

#2Na_3PO_4(aq) + 3Pb(NO_3)_2(aq) rarr Pb_3(PO_4)_2(s)darr + 6NaNO_3(aq)#

#"Moles of lead nitrate"# #=# #(12.4*cancelg)/(331.2*cancel(g)*mol^-1)# #=# #0.0374 *mol#.

#"Moles of trisodium phosphate"# #=# #(17.3*g)/(163.94*g*mol^-1)# #=# #0.106*mol#.

There is a slight excess of lead nitrate: #3xx0.0374*mol" lead nitrate"=0.112*mol#. See if you can work the excess in grams.

Note that if we did the actual experiment, we would probably get #PbHPO_4(s)#... i.e. #"lead(II)biphosphate..."#

How would we write the equation between #Na_3PO_4# and #Pb(NO_3)_2#, given #17.3*g# of the former, and #12.4*g# of the latter?