Question

QUESTION 1 Consider the following reaction: 4FeCr2O4(s) + 8K2CO3(aq) + 7020) 2Fe2O3(s) + 8K2CrO4(aq) + CO2(g) . 4.0 g of FeCr204 and 6.0 g of K2CO3 were reacted in excess O2. What was the limiting reagent? CO2 o K2CO4 0 02 K2CO3 FeCr204 Fe2O3 QUESTION 2 Consider the following reaction: 4FeCr2O4(s) + 8K2CO3(aq) + 702(g) — 2Fe2O3(s) + 8K2Cro4(aq) + CO2(g) 3.8 g of FeCr204 and 2.8 g of K2CO3 were reacted in excess O2 What was the theoretical yield of Fe2O3? QUESTION 3 Consider the following reaction: 4FeCr2O4(s) + 8K2CO3(aq) + 702(g) - 2Fe2O3(s) + 8K2CrO4(aq) + CO2(g) . 4.0 g of FeCr204 and 6.0 g of K2CO3 were reacted in excess 02. 10.68 g of Fe2O3 was obtained, what was the percent yield?

Answer 1

y fels, Qu (s) + 81CQ (aq) + 70, 4) -> 28,0,00 + 8k, Croy lag) + 8(0 (9) Hij of ke Chou Since Of is in exceps therefore it 69 of K₂CO Can't be limiting reapent. molecular weight of fe you = 223 g/mol Molecular weight of ķ(Og = 1383/mol No of Moles of fe Couls) in No. I Moles f lqco = - Weight - 4 = 0.0179/ - Height Molecular weight 23 Molector Weight -0.0434 No. of Moles – 0.0179 0.0044 No of Moles Stoichiometric Efficient 0.0434 ... 8 - 0.0054 Stoichiometric Cofficient Since 0.0044 < 0.0054 Therefore felrou is the limiting Reagent f Reaction.

As we already know that le Gou W is the limiting reagent for Reaction. W e 389 7 februn 8.8 3 J kalo from Reaction Equation We know that corresponding to 4 Moles of le Gryon we will get 2 Moles of Pego. Here No of Moles of fe trou = 3.8 = 0.01704 No of Moles of fez (6) an half of fe Cza (6) 123 No of Moles I kg(s): 0.01704 = 0.00852 height - No. I Moles X Moleculor Mass Weight I k 6) = 0.00852 x 160=1363 Theoratical yield I F z = 1.369

s. gield - Experimental hield Theoraticed yield 7. yield - 0.68 736 Post 1 loo x 7. yield - 50%