Question

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`?

Answer 1

`=15.8944g`

Explanation:

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`