Question

A.) What amount of 1.03M H2SO4 would be required to dissolve 1.83g of iron (steel) wool?

(b) What is the theoretical yield of the Mohr's salt if the above mass of iron is the limiting reagent?

(c) If 11.36 g of Mohr's salt is isolated by vacuum filtration, what is the percent yield of the reaction?

Answer 1

A)

31.81ml

Explanation

The reaction between H2SO4 and Fe is as follows

Fe(s) + H2SO4 (aq) -----> FeSO4(aq) + H2(g)

Stoichiometrically, 1mole of Fe reacting with 1mole of H2SO4

Number of moles = Mass /molar mass

Number of moles of Fe = 1.83g/55.854g/mol = 0.03276mol

so,

Number of moles of H2SO4 required = 0.03276mol

Volume of 1.03M H2SO4 solution containing 0.03276mol of H2SO4 = (1000ml/1.03mol)×0.03276mol = 31.81ml

B)

12.85g

Explanation

1mole of mohr's salt is produced from 1mole of Fe²⁺

No of moles of Fe²⁺ = 0.03276

So,

Number of moles of mohr's salt should be produced = 0.03276

Molar mass of mohr's salt = 392.13g/mol

Mass = Number of moles × molar mass

mass of mohr's salt of should be produced = 0.03276mol ×392.13g/mol = 12.85g

Therefore,

Theoretical yield = 12.85g

C)

88.40%

Explanation

percent yield = (Actual yield/Theoretical yield)×100

Actual yield = 11.36g

Theoretical yield = 12.85g

percent yield = (11.36g/12.85g)×100

percent yield = 88.40%