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