Percent Yield 1. For the reaction of nitrogen and hydrogen to produce ammonia: 3

Question

Percent Yield 1. For the reaction of nitrogen and hydrogen to produce ammonia: 3 Hz (g) + N2 (g) 2 NH3 (g) a. If 10.0 g of Nz were reacted, what is the theoretical yield of ammonia, in grams? b. If 10.0 g of N2 were reacted, and 9.1 g of ammonia were obtained, what is the % yield? 2. Potassium superoxide, KO2, reacts with carbon dioxide to form potassium carbonate and oxygen. This reaction makes potassium superoxide useful in a self-contained breathing apparatus. a. How much oxygen could be produced from 2.50 g of KO2 and 4.50 g of carbon dioxide? b. If only 0.500 g of oxygen is produced from 2.50 g of KO2 and 4.50 g of carbon dioxide, what is the % yield? 3. After 3.8 mL of liquid ethanol (C2HsOH, density-0.789 g/mL) is allowed to burn in the presence of 12.5 g of oxygen, 3.10 mL of water (density 1.00 g/mL) is collected. Determine the limiting reactant, theoretical yield of H20 and percent yield for the reaction. (You will need to write the balanced chemical equation for the combustion of ethanol.) 4. Phosgenite, a lead compound with the formula Pb Cl2cOs, is found in ancient Egyptian cosmetics Phosgenite was prepared by the reaction of PbO, sodium chloride, water, and carbon dioxide; besides phosgenite, the other product formed is aqueous sodium hydroxide. PbO (s) + Nacl (aq) + H2O (/) + CO2 (g) Pbycl2CO3 (s) + NaOH (aq) a. How many grams of phosgenite can be obtained from 10.0 g of PbO and 10.0 g of sodium chloride in the presence of excess water and carbon dioxide. b. If 2.72 grams of phosgenite are produced in the laboratory from the amounts of starting materials stated in (b), what is the percent yield of the reaction?

Explanation / Answer

Solution:- (1) The given balanced equation is...

3H2(g) + N2(g) -----> 2NH3(g)

(a) Grams of N2 ----> moles of N2 ---> moles of NH3 -----> grams of NH3

10.0 g N2 x (1mol N2/28 g N2) x (2 mol NH3/1 mol N2) x (17 g NH3/1 mol NH3) = 12.1 g NH3

(b) Theoretical yield calculated in first part is 12.1 g and the actual yield is given as 9.1 g.

Percent yield = (actual /theoretical)*100

percent yield = (9.1/12.1)*100 = 75%

(2) The balanced equation is...

4KO2 + 2CO2 -----> 2K2CO3 + 3O2

(a) First of all we need to figure out the limiting reactant and for this we can calculate the grams of one reactant for the given grams of another and see if the calculated amount is bigger or less than it's given amount.

If calculated amount is less than avalilable amount then it is excess reactant and if the calculated amount is greater than given amount then it is limiting reactant.

2.50 g KO2 x (1mol KO2/71.1 g KO2) x (2 mol CO2/4 mol KO2) x (44.01 g CO2/ 1mol CO2)

= 0.774 g CO2

Calculated amount is 0.774 g and given one is 4.50 g. It means CO2 is limiting and KO2 is limiting.

2.50 g KO2 x (1mol KO2/71.1 g KO2) x (3 mol O2/ 4 mol KO2) x (32 g O2 / 1 mol O2)

= 0.844 g O2

(b) Percent yield = (0.500 /0.844)*100 = 59%

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