Can some please answer this question for me I don’t understand

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 16.7 g of sulfuric acid and 11.4 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?         Select an answer H₂SO₄ Pb(C₂H₃O₂)₂ HC₂H₃O₂ PbSO₄ there is no 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

H₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) ==> PbSO₄(s + 2 HC₂H₃O₂(aq) ... balanced equation

Since you are given the amounts of BOTH reactants, you must find which one is in limiting supply. An easy way to do this is to divide the MOLES of each reactant by its corresponding coefficient in the balanced equation and whichever value is less identifies the limiting reactant (see below).

For H2SO4: 16.7 g H2SO4 x 1 mol / 98 g = 0.170 moles H2SO4 (÷1->0.17)

For Pb(C₂H₃O₂)₂: 11.4 Pb(C₂H₃O₂)₂ x 1 mol / 325 g = 0.0351 moles Pb(C₂H₃O₂)₂ (÷1->0.035)

Since 0.035 is less than 0.17, lead acetate, Pb(C₂H₃O₂)₂ is the limiting reactant.

Next, the moles of the limiting reactant (0.0351 mols) will determine how many moles of H2SO4 are used and therefore, how many moles / grams of H2SO4 are left over. Since lead acetate is limiting, all of it will be used up and there will be none of left over.

To calculate grams H2SO4 left over ...

0.0351 mols lead acetate x 1 mol H2SO4 / mol lead acetate= 0.0351 mols H2SO4 used

0.170 mols H2SO4 - 0.0351 mols = 0.135 mols H2SO4 left over

mass of H2SO4 left over = 0.135 mols H2SO4 x 98 g / mol = 13.2 g H2SO4 left over

To find grams PbSO4 produced ....

0.0351 mols lead acetate x 1 mol PbSO4 / mol lead acetate x 303 g PbSO4 / mol = 10.6 g PbSO4

To find grams of acetic acid (HC₂H₃O₂) produced:

0.0351 mols lead acetate x 2 mols HC₂H₃O₂ / mol lead acetate x 60.1 g HC₂H₃O₂ /mol = 4.22 g HC₂H₃O₂