# 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?

Dec 28, 2017

$= 15.8944 g$

#### Explanation:

1. Write and balance the equation
$2 P b O \left(s\right) + 2 N a C l \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \to P {b}_{2} C {l}_{2} C {O}_{3} \left(s\right) + 2 N a O H \left(a q\right)$
2. Find the molar masses of the involved compounds
$P b O = \frac{223 g}{m o l}$
$N a C l = \frac{58.5 g}{m o l}$
$P {b}_{2} C {l}_{2} C {O}_{3} = \frac{545.3 g}{m o l}$
3. Given the masses of the involved reactants, per convention, convert it to moles through molar conversions.
$a . P b O$
$= 13.0 \cancel{g P b O} \times \frac{1 m o l P b O}{223 \cancel{g P b O}}$
$= 0.0583 m o l P b O$

$b . N a C l$
$= 13.0 \cancel{g N a C l} \times \frac{1 m o l N a C l}{58.5 \cancel{g N a C l}}$
$= 0.2222 m o l N a C l$
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.
$a . \eta P b O = 0.0583 m o l$
$= 0.0583 \cancel{m o l P b O} \times \frac{2 m o l N a C l}{2 \cancel{m o l P b O}}$
$= 0.0583 m o l N a C l$

$\text{This means that } 0.0583 m o l P b O \equiv 0.0583 m o l N a C l$.

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

$\text{This case, NaCl is the x's reactant}$

$b . \eta N a C l = 0.2222 m o l$
$= 0.2222 m o l N a C l \times \frac{2 m o l P b O}{2 m o l N a C l}$
$= 0.2222 m o l P b O$

$\text{This means that } 0.2222 m o l N a C l \equiv 0.2222 m o l P b O$.

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

$\text{This case, PbO is the limiting reactant}$
5. Since the limiting reactant is already identified, this will be the basis for the determination of the maximum production of $P {b}_{2} C {l}_{2} C {O}_{3}$. 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)
$= 15.8944 g P {b}_{2} C {l}_{2} C {O}_{3}$