Question

If 50.00 mL of 0.200 M ammonium bromide and 25.00 mL of 0.300 M lead(II) nitrate are mixed, what is closest to the theoretical yield of PbBr2? The correct answer is 1.84g but I keep getting 2.75g.

Answer 1

Pb(NO3)2(aq) + 2NH4Br(aq) ----> PbBr2(s) + 2NH4NO3(aq)

25mL of a0.300M Pb(NO3)2 solution would contain

.025L x 0.300 mol/L= 0.0075mol Pb(NO3)2

and

50mL of a 0.200M NH4Br solution would contain

0.05L x 0.200 mol/L=0.01mol NH4Br

NH4Br would be your limiting reagent in this case since it has the least number of moles and will be consumed in the reaction first. for 0.0075 moles of pb(no3)2 it requires 0.015 moles of NH4Br but the reaction contains only 0.01 moles of NH4Br so NH4Br will be the limiting reactant as it is consumed first

It takes 2mol of (looking at the equation above) of NH4Br to make one mole PbBr2, so the number of moles of PbBr2 produced is half of the NH4Br mol.

We need to convert this to grams, so we use the equation:

0.010mol/ 2 x 367g/mol= 1.835g of PbBr2 produced. which is equal to 1.84 grams