When solutions containing 1.00 g of BaCl 2 and 1.00 g of Na2SO4 were mixed, soli

Question

When solutions containing 1.00 g of BaCl2 and 1.00 g of Na2SO4 were mixed, solid BaSO4 formed. The BaSO4 precipitate wasfiltered out of the solution, allowed to dry, and then weighed. Themass of BaSO4 that was isolated from solution was 1.04 g. Use thisinformation to answer the following questions. Pay close attention to the proper use and reporting of significant figures and units.

(a) Calculate the moles of BaCl2 that areavailable for reaction.

(b) Calculate the grams ofBaSO4 that should form if allof the available BaCl2reacts.

(c) Calculate the moles ofNa2SO4 that are available for reaction.

(d) Calculate the grams ofBaSO4 that should form if allof the available Na2SO4reacts.

(e) Which reactant (BaCl2 orNa2SO4) is the limiting reagent in this reaction?Why?

(f) What is the theoretical yield of BaSO4?

(g) What is the percent yieldof BaSO4?

(h) If this reaction went tocompletion, then how many grams of the excess reagent should remainafter the reaction is complete?

Explanation / Answer

a.) BaCl?(aq) + Na?SO?(aq) ? 2 NaCl(aq) + BaSO? (s)
b.) 1.00-g BaCl? x ( 1 mol BaCl? / 204-g BaCl?) = 0.00481-mol BaCl?
c.) 0.00481-mol BaCl? x ( 1 mol BaSO? / 1 mol BaCl?) x ( 233-g BaSO?/ 1mol BaSO?) = 1.12-g BaSO?
d.)1.0-g Na?SO? x (1 mol Na?SO? / 142-g Na?SO?) = 0.00704-mol Na?SO?
e.) 0.00704-mol Na?SO? x ( 1 mol BaSO? / 1 mol Na?SO?) x ( 233-g BaSO?/ 1mol BaSO?) = 1.64-g BaSO?)
f.) BaCl? is limiting reagent because the balanced equation shows equal moles of the reagents are needed. Since we have only 0.00481 moles of BaCl? and 0.00704 moles of Na?SO?, the smaller quantity of BaCl? makes it the limiting reagent.
g.) Theoretical yield of BaSO? is 1.12-g (see answer 'c')
h.) % yield = actual yield / theoretical yield x 100% = 1.04-g / 1.12-g x 100% = 92.9%
i.) equal moles of reactants are needed, so if we use 0.00481-mol of each, we should have 0.00704 - 0.00481 = 0.00233-mol Na?SO? un-reacted. That would be
0.00233-mol Na?SO? x ( 142-g Na?SO? / 1 mol Na?SO?) = 0.317-g Na?SO?

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