1)If 0.238 moles of zinc reacts with excess lead(IV) sulfate, how many grams of
ID: 540117 • Letter: 1
Question
1)If 0.238 moles of zinc reacts with excess lead(IV) sulfate, how many grams of zinc sulfate would be produced in the following reaction?
Pb(SO4) + 2Zn -> 2ZnSO4 + Pb
2)If a solution containing 55 g of mercury(II) nitrate is allowed to react completely with a solution containing 14.334 g of sodium sulfate according to the equation below:How many grams of solid precipitate will be formed?How many grams of the reactant in excess will remain after the reaction?
3)In the following reaction, 451.4 g of lead reacts with excess oxygen forming 337.6 g of lead(II) oxide. Calculate the percent yield of the reaction.
2Pb(s)+O2(g) ------> 2pbO(s)
Explanation / Answer
1) Pb(SO4) + 2Zn -> 2ZnSO4 + Pb
1 mol Pb(SO4) = 2 mol Zn = 2 mol ZnSO4
no of mol of ZnSO4 produced = 0.238 *2/2 = 0.238 mol
Amount of ZnSO4 FORMED = 0.238* 161.44 = 38.42 g
2) Hg(NO3)2 (aq) + Na2SO4(aq) ---> HgSO4(s) + 2NaNo3(aq)
1 mol Hg(NO3)2 = 1 mol Na2SO4
No of mol of Hg(NO3)2 = W/MWT = 55/324.6 = 0.17 mol
No of mol of Na2SO4 = W/MWT = 14.334/142.04 = 0.1 mol
limiting reactant = Na2SO4
No of mol of HgSO4 formed = 0.1 mol
Amount of precipitate = 0.1*296.653
= 29.66 g
amount of excess Hg(NO3)2 = (0.17-0.1)*324.6 = 22.722 g
3) 2Pb(s)+O2(g) ------> 2pbO(s)
no of mole of Pb = w/mwt = 451.4/207.2 = 2.18 mol
theoretica yield of PbO = 2.18*223.2 = 486.576 g
percent yield = practical yield/theoretical yield*100
= 337.6/486.576*100
= 69.4%
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