Can someone please help me with this problem????

Question

H2SO4(aq) + Pb(C2H3O2)2(aq) --> PbSO4(s) + 2 HC2H3O2(aq)

Solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and a solution of acetic acid. If 18.6 g of sulfuric acid and 12.6 g of lead (II) acetate are mixed, calculate the number of grams of sulfuric acid,

lead (II) acetate, lead (II) sulfate, and acetic acid present in the mixture after the reaction is complete.

Which chemical is the limiting reactant?  

How many grams of sulfuric acid are leftover? grams

How many grams of lead (II) acetate are leftover? grams

How many grams of lead (II) sulfate are produced? grams

How many grams of acetic acid are produced? grams

Answer 1

I suppose, given the information of the problem, that we are not considering equilibrium constants. In that case:

1) Which chemical is the limiting reactant? Well, in our balanced equation, we need a 1:1 mole ratio. Therefore, whichever component has fewer moles is the limiting reactant.

Sulfuric acid: 18.6 g / 98.079 g/mol = 0.1896 moles

Lead acetate: 12.6 g / 325.29 g/mol = 0.0387 moles

It is clear that the lead acetate is the limiting reactant.

2) How many g of sulfuric acid are left? We know that we're using 0.0387 moles of each reactant based on question 1. Therefore, we use ALL of our lead acetate and are left with 0.1896 - 0.0387 = 0.1509 moles sulfuric acid. We convert to g using stoichiometry.

0.1509 moles * 98.079 g/mol =14.8 g unreacted sulfuric acid.

3) How many grams of lead (II) acetate are leftover? This was answered in question 2. Since it's the limiting reagent, we used all of it, so 0g are left.

4) How many grams of lead (II) sulfate are produced? In the balanced reaction, it also has a molar ratio of 1. Therefore, we used 0.0387 moles of reactants and produced 0.0387 moles of lead sulfate. Calculate the number of g using stoichiometry:

0.0387 mol * 303.26 g/mol = 11.74 g lead (II) sulfate

5) How many grams of acetic acid are produced? Just like in question 4, we must be aware of molar ratios. We know that 0.0387 is one molar ratio. In the balanced equation, though, we produce TWO molar ratios of acetic acid. We simply multiply by 2 to get the number of moles of acetic acid produced:

0.0387 * 2 = 0.0774 moles acetic acid.

Convert moles to g using stoichiometry:

0.0774 mol * 60.052 g/mol = 4.65 g acetic acid