General Chemistry Final Exam Review Problems Score: Name: The Rules o. Answers m

Question

General Chemistry Final Exam Review Problems Score: Name: The Rules o. Answers must be placed in the space provlded at the conclusion of the problem b. Al work must be shown on paper you attached to the document Your method of solution needs to be detoiled so that you show demonstrotive ewidence of your ablity to solve the problem c. Course notes and your textbook are valuable sources of information in solving the problems work is your own. I do not expect you te copy from one another. Evidence of this type of behavior wl result in an F for the assignment and an investigation by the Judicle 1. The compound Asale is synthesized by the reaction of arsenic metal with arsenic trilodide.If a solid block of arsenic (de 5.72 g/cm") that is 3.00 cm on edge is allowed to react with 1.01 x 1024 molecules of arsenic triodde, how much Asale can be prepared? If the percent yield of Asle was 75.6%, what mass of Assla was actually isolated? Answer 2. Tris(pentafluorphenyl)borane, commonly known by its acronym BARF, Is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. it is composed of solely ofc, F and B; it is 42.23% C and SS.66% F by mass, what is the empirical formula of BARF? A 2251 E sample of BARF dissolved in 347.0 ml of solution produces a 0.01267 M solution. What is the molecular formula of BARF? Answer 3. The unknown acid HoX can be neutralized completely by OH- according to the following (unbalonced) equation The ion formed as the product, X, was shown to have 36 total electrons. What is the element x? Propose a name for the acid, HaX. To completely neutralize a sample of HuX, 35.6 ml of 0.175 M OH-solution was required. What was the mass of HX sample used? Answer

Explanation / Answer

1. You want As2I4 from As and AsI3 so:

As + AsI3 ===== As2I4 , the balanced equation is

2As + 4AsI3 ===== 3As2I4

The statement says that you have a solid block with 3 cm of edge, we will assume a cube , volume for a cube is

33 = 27cm3

density = mass / volume

mass = density * volume

mass of Arsenic = 5.72 g/cm3 * 27 cm3 = 154.44 grams of As; molar mass of arsenic

moles of Arsenic = 154.44 / 75 = 2.0592 moles of As

For AsI3 we have been provided with 1.01 x 1024 molecules , we know that 1 mole has 6.022 x 1023 so:

1.01 x 1024 / 6.022 x 1023 = 1.68 moles of AsI3

2 moles of As needs 4 moles of AsI3 and we have a smaller value than this so AsI3 is the limiting reactant, we must make calculation according to this value

moles of As2I4 that can be produced:

3 * 1.68 / 4 = 1.26 moles of As2I4

molar mass of As2I4 is 657.5 g/gmol

mass = moles * molar mass = 1.26 * 657.5 = 828.45 grams this is the theoretical yield

for the actual yield we just need to multiply the last value by the actual yield

828.46 * 0.756 = 626.315 grams of product isolated.

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