posted by paul .
8.76g of aluminum sulphide is added to 0.0350L of 0.250mol/L lead(II)nitrate solution. calculate the maximum mass of precipitate, PbS that can form
Al2S3 + 3Pb(NO3)2 ==>3PbS + 2Al(NO3)3
Convert 8.76 g Al2S3 to moles. moles = grams/molar mass
Convert the Pb(NO3)2 to mols. moles = M x L.
Now, using the coefficients in the balanced equation, convert each of the reactants to moles of PbS. Undoubtedly you will obtain different answers for moles PbS; the correct answer in limiting reagent problems like this is the smaller value. Then convert the smaller value to grams. g = moles x molar mass.