CHEM 101A Lecture 10: chem week 10 lecture

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chocolateferret587

3 Nov 2016

School

University of Arizona

Department

Chemistry

Course

CHEM 101A

Professor

Laura Van Dorn

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Document Summary

Amount of product obtained when the reaction takes place. Many times in chemical reaction there is not enough of one product of one reactant to use up the other reactant completely. Limits amount of product you can make. Once product is starting to be made, products will start making reactant. Reversible reaction can be approached from either direction. After certain amount of time, concentrations of reactants and products are constant. Forward and reverse reactions continue after equilibrium. K subscript eq gives concentrations of products and reactants at equilibrium. No matter what concentrations you start with, at equilibrium, the ratio of products over reactions is constant. Solids and pure liquids do not change their concentration at equilibrium and are not included in the keq expression. A reaction that favors products has a large k. A reaction with about equal concentrations of products and reactants has a k close to one. A reactions that favors reactants has a small k.

Related Questions

I need help on number 7!!!

Balance the following equation, using simplest, whole number coefficients. Then answer the following by using the balanced equation, considering each part independent of one another.

___ Zn(s) + ___ Pb(NO3)4 (aq) --> ___ Pb(s) + ___ Zn (NO3) 2 (aq)

a) How many moles of zinc would be needed to yield 7.25 moles of zinc nitrate?

b) how many grams of lead(IV) nitrate will produce 1.47g of zinc nitrate ?

c) What mass of lead would be formed when 75.0g of Zn and 25.0g of Pb(NO3)4 are reacted? Identify the limiting and excess reactant (by formula).

limiting reactant _______________ excess reactant____________________

If the reaction in part C above was performed in the lab with those starting quantities of reactants and ended up with an 81 % yield, how much Pb was actually isolated in the lab?

1. Potassium aluminum sulfate dodecahydrate, KAl(SO₄)₂·12H₂O, is a member of the alum salts. It has a molar mass of 474 g/mol and beautiful solid crystals can be grown from an aqueous solution according to this chemical equation:

The next three questions are based on a synthesis of the alum salt product performed by Kemmi Major. To begin, Kemmi weighs 0.50 g of aluminum metal. She dissolves the aluminum with a strong base to obtain Al³⁺ cations and adds an excess of the other reactant ions (K⁺ and SO₄^2-) to the solution. She cools the solution in an ice bath until crystals of alum salt form. What is the theoretical yield of alum product in grams if the reaction works perfectly? (Hint: convert the limiting reactant aluminum to moles of Al. Assume each Al atom gives one Al³⁺ cation. Use the balanced chemical equation to predict the moles of alum product, and convert this to grams using the molar mass given.) Round the answer to 2 significant digits, with units of g.

2. Kemmi dries her alum crystals and weighs them. Her experimental mass is 6.7 g of alum product from her experiment. What is Kemmi’s percent yield? (Remember, the answer to the previous question is her theoretical yield.) Report 2 significant digits with units of %.

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CHEM 101A Lecture 10: chem week 10 lecture

Document Summary

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