Answer to Question #179934 in General Chemistry for Ally

Question

In a lab, a student mixes 25.0 ml of 0.115 M aluminum sulfate with 34.0 ml of 0.0975 M Lead (III) nitrate. Calculate the grams of the precipitate that is formed.

Answer 1

Molar masses

Al2(SO4)3= 342g/mol

Pb(NO3)2=331g/mol

PbSO4=303g/mol

Let's find the masses of each reactant.

Mole= concentration x volume

Mole of Al2(SO4)2= 0.115x25/1000=0.002mol

Mass of Al2(SO4)2= mole x molar mass

= 0.003x342=0.98g

Also,

Mole of Pb(NO3)2= 0.0975x34/1000= 0.0033mol

Mass= 0.0033x331=1.01g

Now, let's find the limiting reagent that determines the mass of precipitate from the balanced equation.

Al2(SO4)3 + 3Pb(NO3)2 ---> 3PbSO4 + 2Al(NO3)3

342g of Al2(SO4)3 reacts with 3(331)g of Pb(NO3)2

0.98g of Al2(SO4)3 should react with 331x0.98/342 = 2.85g of Pb(NO3)2

But only 1.01g of Pb(NO3)2 is available, therefore, it is the limiting reagent.

Now,

331g of Pb(NO3)2 gives 303g of PbSO4

1.01g of Pb(NO3)2 will give 303/331 x 1.01 = 1.007g of PbSO4

Therefore, the mass of precipitate PbSO4 produced is 1.007g