Question

Percent yield of chemical reactions Gaseous butane (CH(CH), CH) reacts with gaseous oxygen gas (02) to produce gastous carbon dioxide (CO2) and gaseous water (H,0). IF 10.8 g of water is produced from the reaction of 14.5 g of butane and 68.1 g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures. X $ ?
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Answer 1

From the reaction Gaseous butane (Cų (CH2) CH) = Ho (9) haseous oxygen (3)= 319) The reaction is C4 Mo (9) + 819) - cqt the The balanced equation Ci Ho (9) +6-502 (9) -->400 (9)+ 540 (9) 244 Mio (9) + 13019) -581819) + 10420 (g). From the balanced equation a mole Ca Ho reacts with 13 molo The mole ratio of Ca Ho and of is C4 HD = @= 2:13= 1:6.5 From the data 14.5g of butane and 68.18 of a number of modes = weight molay mass butane C4 Mio molar mass = 1214) + 10(1) = 589/mol 14.59 = 589/mo = 0.25 mol Scanned with CamScanner

©2 Oz molar mass = Q(16) = 3ag/mol n= 68. 19 2:13 mol 32g/mol The mole ratio is Gu Ho: a = 0.25: 2:13 - 025 ; 2.13 0.25 0.25 = 1:8.52 From balanced eq" : GMO i = 1:6.5 From given data : cho 02 = 1:8.5 Excessai is present. Limiting reagent is Cato product quantity depends on the from balanced equation a mole cu ho produce 10 ml hon 2 mol Cy Ho - lomol tho 1 mol Cu Ho - 5 mol the - 589 Catho - 5 (189) of the 14.59 C4 Mo → ? 5 (189) of the va 2 x (14 5g CL Hio) = 22:59 hol 589 CAMO Scanned with cam Scanner

.: Theoretical Yield of HO = 22.5g. Actual field of tho=10.89 Percent Yield - Actual yield - X100% Theoretical Yield 100% % yield = 10.89 -x 22:59 = 4866 % yield = 48.0% Scanned with CamScanner