Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sul

Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid.5.00g of sulfuric acid and 5.00g of lead(II) acetate are mixed:

A)Calculate the number of grams of sulfuric acid present in the mixture after the reaction is complete.

B)Calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete.

C)Calculate the number of grams of lead(II) sulfate present in the mixture after the reaction is complete.

D)Calculate the number of grams of acetic acid present in the mixture after the reaction is complete.

Explanation / Answer

The balanced equation for given reaction is given by:

H2SO4 + Pb(C2H3O2)2 ? 2C2H4O2 + PbSO4(s)

Molecular weight of sulphuric acid= 98.079g

Molecular weight of lead (II) acetate (Pb(C2H3O2)2 )= 325.29g

molecular weight of acetic acid= 60.05g

Molecular weight of lead(II)sulphate= 303.26g

so, now,

A)As we see in the balanced equation one mole(98.079g)of H2SO4 reacts with one mole(325.29g) of lead acetate.

But we have 5g of both, when we need much higher amount of lead acetate to react completely with 5g of H2SO4

So, clearly here lead acetate is limiting reagent

Now, since 325.29g of lead actate reacts with 98.079g of H2SO4

so, 5g of lead acetate will react with [(98.079/325.29)*5]g of H2SO4= 1.5076g

So, only 1.5076g of H2SO4 will react

so, left over H2SO4 = 5-1.5076= 3.4924g

B) as we see in the above question, lead acetate is the limiting reagent and H2SO4 is present in excess.

Hence, complete 5g of lead acetate will be consume, and hence the amount left at the end will be 0g

C) From the balanced equation we can say that

One mole of lead acetate(325.29g) reacts to form one mole of lead sulfate(303.26g)

and now since we have only 5g of lead acetate, so it will react to form (303.26/325.29)*5= 4.66g lead sufate

D) similarly, from balance equation we can say that

One mole of lead acetate(325.29g) reacts to form two mole of acetic acid(2*60.05= 120.1g)

and again since we have only 5g of lead acetate, so it will react to form (120.1/325.29)*5= 1.846g acetic acid

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