3.29 Question 2
of iron is added to 50cm3 of 1.0 mol/dm sulphuric acid,
Fe + H2SO4 FeSO4+H2
(a) Calculate the number of moles of sulphuric acid present,
(b) Show by calculation that sulphuric acid is the limiting reactant. [2]
(c) Calculate the mass of iron remaining after the reaction is complete. [2]
(d) Calculate the maximum volume of hydrogen formed in the reaction at rt. p [2]
(e) Calculate the theoretical yield of iron(II) sulphate at the end of the reaction. [2]
(1) The actual yield of iron(II) sulphate was 3.61 g. Calculate the percentage yield.​

The correct answer is "false"

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.