Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
[tex]= \frac{Mass}{Molar\ mass}[/tex]
[tex]= \frac{10.0 g}{259.65g / mol}[/tex]
= 0.0385 mol
Mass of PbO needed is
[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.