Calculate the mass of lead bromide that could be made when 154. mL of 0.450 M lead nitrate reacts with excess sodium bromide according to the following reaction:

Question

Pb(NO3)2(aq) +. 2NaBr(aq). --->. PbBr2(s). +. NaNO3(aq)

Answer 1

This a stoichiometry problem. Your key skills here would be:

(a)  Identifying the limiting reagent. In this case, it is lead nitrate because you were told that sodium bromide is in excess.

(b)  Converting from moles to grams in order to obtain the mass of the final product.

First calculate the moles of Pb(NO₃)₂ you are starting with. Use the formula:

Moles = molarity x volume. Watch your units – 154. mL must first be converted to liters.

(0.154 L)(0.450 mol/L) = 0.0693 mol of Pb(NO₃)₂. This is also equal to moles of Pb and in the presence of excess NaBr, will also be equal to moles of PbBr₂ formed. Look at the coefficients of these substances in the balanced equation to determine the mole ratio of lead nitrate to lead bromide (1:1).

Now convert the moles of PbBr₂ formed to mass of PbBr₂. The conversion factor is the molar mass of PbBr₂, 367 g/mol.

Mass of PbBr₂ formed = (0.0693 mol)(367 g/mol) = 25.4 g PbBr₂

It’s always advisable to include units in all your calculations so you can use dimensional analysis to determine if your answer is right.