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#?

Question

Serenity Allen · Accepted Answer

#=15.8944g#

#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#

Wyatt Atkins · Answer

undefined To determine the grams of phosgenite produced, you need to first find the limiting reactant between PbO and NaCl. Then, calculate the moles of the limiting reactant, use the stoichiometry of the balanced equation to find the moles of phosgenite produced, and finally convert moles to grams using the molar mass of phosgenite. The balanced equation for the reaction is:

PbO + 2 NaCl + 2 CO2 + 2 H2O → PbCl2 + Na2CO3 + CO2

The molar mass of PbO is 223.2 g/mol, NaCl is 58.44 g/mol, and phosgenite (PbCl2) is 278.1 g/mol.