Question

Question#1:

Aspirin is prepared by reaction of salicylic acid (C7H6O3)(C7H6O3) with acetic anhydride (C4H6O3)(C4H6O3) according to the following equation:
C7H6O3Salicylicacid+C4H6O3Aceticanhydride→C7H6O3Salicylicacid+C4H6O3Aceticanhydride→ C9H8O4Aspirin+CH3COOHAceticacid

A) How many grams of acetic anhydride are needed to react with 7.50 gg of salicylic acid?

B)How many grams of aspirin will result?

C) How many grams of acetic acid are formed as a by-product?

Question #2:

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)2Al(s)+3Cl2(g)→2AlCl3(s)

You are given 26.0 gg of aluminum and 31.0 gg of chlorine gas.

A) If you had excess chlorine, how many moles of of aluminum chloride could be produced from 26.0 gg of aluminum?

Express your answer to three significant figures and include the appropriate units.

B) If you had excess aluminum, how many moles of aluminum chloride could be produced from 31.0 gg of chlorine gas, Cl2Cl2?

Express your answer to three significant figures and include the appropriate units.

Question #3:

Equation= 3H2(g)+N2(g)→2NH3(g) 1.52 gg H2H2 is allowed to react with 9.81 gg N2N2, producing 1.18 gg NH3NH3.

A)What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

B) What is the percent yield for this reaction under the given conditions?Express your answer to three significant figures and include the appropriate units.

Question #4:

Ethyl alcohol is prepared industrially by the reaction of ethylene, C2H4C2H4, with water.

A) What is the percent yield of the reaction if 4.7 gg of ethylene gives 4.8 gg of ethyl alcohol?
C2H4Ethylene(g)+H2O(l)→C2H6OEthylalcohol(l)C2H4Ethylene⁡(g)+H2O(l)→C2H6OEthylalcohol⁡(l)

Express your answer using two significant figures.

Question #5:

Hydrogen and chlorine react to yield hydrogen chloride:
H2+Cl2→2HCl.

A) How many grams of HClHCl are formed from reaction of 3.58 gg of H2H2 with 8.96 gg of Cl2Cl2?

B) Which reactant is limiting?

Question #6:

How many grams of the dry-cleaning solvent 1,2-dichloroethane (also called ethylene chloride), C2H4Cl2C2H4Cl2, (Figure 1) can be prepared by reaction of 15.4 gg of ethylene, C2H4C2H4, with 3.64 gg of Cl2Cl2? Answer in grams
C2H4+Cl2→C2H4Cl2

Answer 1

Answers) The readion of salicylic acid and acetic anhydride is as follows : – Cy HG 03 (salicylic acid) + CH 03 (neclic anhydride) -> Cghg ºf Caspirin) + CH3COOH (acerebuid) --- A) We know, 0.0543 moles Imole 11 11 た mass of the substance No of moles of a substance Molar mass of the substance ". No of moles of salicylic acid a mass of salicylic aud Molar mass of salicydic acid 7.50 g 138.121 g/mol From equation we can see that Salicylic acid reacts with Imole acetic anhydride 0.0543 (10.0543) 0.0543 mole Mass of acetic anhydride = No of moles of acetic anhydride x Molar mass of acelic anhydride 0.0543 mgle X 102.09 g Imole 5.543 3 g :: Mass of acetic anhydride is g. (Ansurer) I B) from eduation 0 we can see that I mole salicylic acid gives 1 mole of aspirin 0.0543 (1 x 0.0543) = 0.0543 mole Mass of aspirin = no of moles of aspirin x Molar mass of aspirin 0.0543 mole x 180.158 g Imoler 9.782 .. 9.782 g of aspirin will result. (Answers 7 c) From equation o we can see that, 1 mole salicylic acid gives 1 mole acetic aid 0.0543 1 (170.0543) -0.0543 mole ... Mass of acetic acid = No of moles of acetic and & Molar mass of acetic acid 0.0543 mble x 60052 3:26 5.543 g 1/ こ g た g Image ... 3.26 g of acetic acid are formed as a by product. Candures) فه