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?

Answer 1

#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..."#

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Answer 2

Luna Chen

2 Na3PO4 + 3 Pb(NO3)2 → 6 NaNO3 + Pb3(PO4)2

To find the limiting reactant and determine the amount of the excess reactant remaining, you need to compare the moles of each reactant using their respective molar masses.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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?

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?