Question

Please answer what you can, or are willing to. Thanks so much!!

1. You have develop a synthesis to produce glucose (C6H12O6) from carbon monoxide and hydrogen peroxide, the by produce of the reaction is oxygen gas. The reactions has a average yield of 78.2%. If you want to bubble Carbon monoxide through the hydrogen peroxide solution and obtain 67.4g of glucose, how much carbon monoxide would you need to use?

2. Suppose the theoretical yield for a reaction is 19.87g and your yield is 14.79g, what is the percent yield of your reaction?

3.  Determine the molar concentration of a solution that was make by dissolving 45.6g of Ba(NO3)2 in 2.32L of solution

4.  A piece of Al is dropped in a large flash of concentrated hydrochloric acid, the gas that come off is collected by water displacement. 256.1 mL of gas is collected by water displacement at 20.0o C with a total atmospheric pressure of 763mmHg. If you assume a 100% yield how big (in grams) was the piece of Al?

5.  In a titration experiment it was determine 34.56mL of a KOH solution was needed to reach the equivalence point of a 78.0mL

6. Determine the liters of Oxygen need for the complete combustion of 768g of benzene (C6H6) if the reaction is run at 23.5o C and 765mHg?

Answer 1

page-t Q.1 co + H₂O Color of t O₂ The reaction has a average yield of 78.2%. If you want to bubble Carbon monoxide (col through the hydrogen peroxide solution and obtain 67.4 g of glucose how much Carbon monoxide would you need to use? son - +60 6 co + 6H₂O Carbon monoxide =28.01 g/mol Chez oo glucose molar mals 180.156 g/mol The theoretical yield wee assume as a given average yield = 78.2 % and obtained yield = 67.4 g ( glucose) so theoretical yield a s 69.4 x 100 78.2 = 86.189 g ( glucose) from the balanced chemical equation, t mole of glucose is formed from 6 mote of co 180.156 g of glucose is formed from 6x 28.01 g of - | 6 8 06 3 (com Alene, Bb. 189 g (glucosel is formed from 168°06 x 180.156 or and 180.40 g of carbon monoxide é 80.40 g of (col = 2.870 mol of cool

2 page-2 Theoretic obtained yield (given) = 19.87 g yield = 14.799 Perunt yield obtained yield x doo % & theoretical gield 14.79 oxdoo / 19.87 I Percent gield 44.43 % L 3 Criven , mass of Ba(NO3)2 = 48.6 g volume of Seth - 2.32 L now, molar mass of Balnog = 261.337 g/mol If the molar concentration is consider IM, then 1 lithe solution contains I mole Ba No. 261.337 g Form so 2.32 L solution contains = 1261.337 X 20327 g = 606.301 g now, 606.301 g Balogh means I m sotation so 45.6 g (Ba(N)) means t x45.6 6061301 - 0.075 M Hence molar Conem of solu - 0.075 m