Question

(A) For the following reaction, 6.86 grams of sulfur dioxide are mixed with excess oxygen gas . The reaction yields 6.48 grams of sulfur trioxide .

sulfur dioxide ( g ) + oxygen ( g ) sulfur trioxide ( g )

What is the theoretical yield of sulfur trioxide ? grams

What is the percent yield for this reaction ? %

----------------------------------------------

(B) For the following reaction, 8.25 grams of nitrogen gas are allowed to react with 1.12 grams of hydrogen gas.

nitrogen (g) + hydrogen (g) ammonia (g)

What is the maximum amount of ammonia that can be formed? grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

(A The kalaned reaction is 2808 (9) + O2(g) 280 CD) Mol. Wt 2x64069 329 2X8006 g Here soo is the limiting exactant from the ahove reaction, 2X64*069 502 prochicas sog = 2*80'06 g .: 6'86 9 " " " = (2x 80'06 X 686)7 - 8'54 g 2x 64 '06 Theoretical yield = 857g y yield = situal field 100% & 686 x 100 % 280005% Theorchical field 8:54 (③) The balanced reaction is N2 (9) + 3H2(9) ► 2013 (9) Malawt: 28 g 322 0169 2x17 g to react completely 28 g of NH3 requires Ma = 8x2'016 g ...825 g " - - 3 X2'016 xE23 g = 14829 Hence, amount of Hy is in less amount than required to react completely. Heme, He is the limiting reactant. 3x2.016 g of H₂ produces NH3 & 2 X 14 g 1.12 g "" " " = 2x17x112 g = 6*296 g Maximum amount of NH₃ can be formed = 6*296 g 3x2 016 g of H2 reacts with N2 = 289 :1/12 g of " 28X112 7 = 5'1859 Therefore, Ny is in exuss amount by = (8.25 - 5 19) g = 3:069 3x2.016 3x 2016