Answer:
The mass of lead(ll)tetraoxosuplhate formed when 0.5mol-¹ of tetraoxosulphate react with lead(ll)chloride is :
151.63 grams
Step-by-step explanation:
PbCl2 = lead(ll)chloride
PbSO4 = lead(ll)tetraoxosuplhate
H2SO4 = hydrogen tetraoxosulphate
HCl = Hydrochloric acid
The balanced equation for the reaction is :
![PbCl_(2)+H_(2)SO_(4)\rightarrow PbSO_(4)+2HCl](https://img.qammunity.org/2021/formulas/chemistry/middle-school/9tv5bn0f7rx67qf6owgh8jkh3ptcjhmw78.png)
The stoichiometric coefficients indicates that :
1 mole PbCl2 = 1 mole H2SO4 = 1 mole PbSO4 = 2 mole HCl
1 mole H2SO4 produces = 1 mole PbSO4
0.5 mole produce = 0.5 mole of PbSO4
We have 0.5 mole of PbSO4
So , we are asked to calculate the mass of 0.5 mole pf PbSO4
Molar mass of PbSO4 = mass of Pb + Mass of S + 4(mass of O)
= 207.2 + 32.06 +4(16)
=303.26 grams
![mass = Molar\ mass* Moles](https://img.qammunity.org/2021/formulas/chemistry/middle-school/a7vwfn4tidawcp2q4kswvdh5ompoxdsm3v.png)
![mass = 303.26* 0.5](https://img.qammunity.org/2021/formulas/chemistry/middle-school/r4ar1lyc08bxztt3l4xptq6j849ipdvypu.png)
151.63 grams