How many grams of phosgenite can be obtained from 13.0 g of #PbO# and 13.0 g #NaCl# in the presence of excess water and #CO_2#?

1. Write and balance the equation
   #2PbO(s)+2NaCl(aq)+H_2O(l)+CO_2(g)->Pb_2Cl_2CO_3(s)+2NaOH(aq)#
2. Find the molar masses of the involved compounds
   #PbO=(223g)/(mol)#
   #NaCl=(58.5g)/(mol)#
   #Pb_2Cl_2CO_3=(545.3g)/(mol)#
3. Given the masses of the involved reactants, per convention, convert it to moles through molar conversions.
   #a. PbO#
   #=13.0cancel(gPbO)xx(1molPbO)/(223cancel(gPbO))#
   #=0.0583molPbO#

#b. NaCl#
#=13.0cancel(gNaCl)xx(1molNaCl)/(58.5cancel(gNaCl))#
#=0.2222molNaCl#
4. Find the limiting reactant. Pair each reactant with each other with reference to the balanced equation to determine the limiting and the x's reactants.
#a. etaPbO=0.0583mol#
#=0.0583cancel(molPbO)xx(2molNaCl)/(2cancel(molPbO))#
#=0.0583molNaCl#

#"This means that " 0.0583molPbO-=0.0583molNaCl#.

#(etaNaCl " available")/(=0.2222molNaCl) > (etaNaCl " required")/(=0.0583molNaCl)#

#"This case, NaCl is the x's reactant"#

#b. etaNaCl=0.2222mol#
#=0.2222molNaClxx(2molPbO)/(2molNaCl)#
#=0.2222molPbO#

#"This means that " 0.2222molNaCl-=0.2222molPbO#.

#(etaPbO " available")/(=0.0583molPbO) < (etaPbO " required")/(0.2222molPbO)#

#"This case, PbO is the limiting reactant"#
5. Since the limiting reactant is already identified, this will be the basis for the determination of the maximum production of #Pb_2Cl_2CO_3#. Refer to the molar masses for the possible conversion factors; thus,
#=0.0583cancel(molPbO)xx(1cancel(molPb_2Cl_2CO_3O))/(2cancel(molPbO))xx(545.5gPb_2Cl_2CO_3)/(1cancel(molPb_2Cl_2CO_3)#
#=15.8944gPb_2Cl_2CO_3#