Consider the following reaction: 2 Fe (NO_3) 3 (aq) + 3 Na_2 S (aq) rightarrow F
ID: 529261 • Letter: C
Question
Consider the following reaction: 2 Fe (NO_3) 3 (aq) + 3 Na_2 S (aq) rightarrow Fe_2S_3(s) + 6 NaNO_3 (aq) a) What type of chemical reaction is it (classification)? b) How many moles of Fe (NO_3) 3 are needed to react with 0.549 mol of Na_2s? c) If 0.500 g of Fe(NO_3) 3 and 0.500 g of Na_2 S are allowed to react in solution, how many grams of Fe_2 S_3 will be produced, and how many grams of excess reactant remain? d) If 2.00 g of Na_2 S reacts with an excess of the other reactant, and 1.45 g of Fe_2 S_3 is recovered, what is the percent yield of the reaction?Explanation / Answer
(7)
(a) It is a double displacement reaction. Because the anions exchange their cations.
(b) From the balanced equation,
3 mol of Na2S needs 2 mol of Fe(NO3)3
then, 0.549 mol of Na2S needs 0.549 * 2 / 3 = 0.366 mol of Fe(NO3)3
(c)
Moles of Fe(NO3)3 = mass / molar mass = 0.500 / 242 = 0.00207 mol
Moles of Na2S = 0.500 / 78 = 0.00641 mol
From the balanced equation,
2 mol Fe(NO3)3 needs 3 mol Na2S
the, 0.00207 mol Fe(NO3)3 needs 0.00207 * 3 / 2 = 0.00310 mol of Na2S ( < 0.00641)
Hence Fe(NO3)3 is limiting reagent.
From the balanced equation,
2 mol Fe(NO3)2 forms 1 mol Fe2O3
then, 0.00207 mol Fe(NO3)3 forms 0.00207 * 1 / 2 = 0.001035 mol of Fe2O3
Therefore, Mass of Fe2O3 = moles * molar mass = 0.001035 * 160 = 0.166 g.
(d)
Moles of Na2S = 2.00 / 78 = 0.0256 mol
From the balanced equation, 3 mol of Na2S = 1 mol of Fe2O3
then, 0.0256 mol of Na2S = 0.0256 / 3 = 0.00855 mol of Fe2O3
Therefore, theoretical yield = 0.00855 * 160 = 1.37 g.
But actual yield = 1.45 g.
% yield = (actual yield / theoretical yield) * 100 = (1.37 / 1.45) * 100 = 94.5 %