# Question #24a44

Apr 27, 2015

To solve a problem like this, first determine the limiting reagent, then calculate the number of moles of product, and finally convert to grams of product.

1) Write the balanced reaction so you'll know the proper stoichiometric ratios. In this case it's easy, but other reactions may be tricky.

$P b {\left(C {H}_{3} C O O\right)}_{2} + {H}_{2} S {O}_{4} \rightarrow P b S {O}_{4} + 2 C {H}_{3} C O O H$

2) Determine which reagent is limiting. In this case there is a 1:1 mole ration of reactants, so we simply want to know which has the smallest number of moles. However, in other cases the mole ratio may be 2:1 or 3:2, etc. so the calculation could be more complicated.

$P b {\left(C {H}_{3} C O O\right)}_{2}$: $\frac{7.20 g}{327.3 \frac{g}{m o l}} = 2.200 \times {10}^{- 2} m o l$
${H}_{2} S {O}_{4}$: $\frac{7.20 g}{98.08 \frac{g}{m o l}} = 7.341 \times {10}^{- 2} m o l$

Lead acetate is the limiting reagent and we expect to form $2.200 \times {10}^{- 2} m o l$ of lead sulfate, assuming 100% yield. Again, this is a simple case because there is a 1:1 ratio of lead acetate and lead sulfate in the balanced reaction.

3) Calculate the number of grams of lead sulfate product.
$P b S {O}_{4}$: $\left(2.200 \times {10}^{- 2} m o l\right) \left(303.26 \frac{g}{m o l}\right) = 6.671 g$