Chegg.com Error: Session Expired X Bb Home Page 15 FALL Texas A&M; Universit nin

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Chegg.com Error: Session Expired X Bb Home Page 15 FALL Texas A&M; Universit ning.com, 2001904 Apps Med Tech sites work Gaming School Downloads Lightroom Tips Archi B solutions sapling learning Sapling Learning Texas A&M; University at Galveston CHEM Lab Fall15 TOWNSEND Activities and Due Dates Balloon Races My Assignment 10/22/2015 02:00 PM A 72.4/100 0/21/2015 12:59 PM Gradebook Attempts score Periodic Table Print Calculator 00 Questi 0 of 10 Map 95 enera emIS presented by Sapling Lea ld McQuarrie .Peter ARock.Etha 90 If a solution containing 121.73 g of silver chlorate is allowed to react completely with a solution containing 10.19 g off lithium hydroxide, how many grams of solid precipitate will be formed? 93 48 How many grams of the reactant in excess will remain after the reaction? 84 00 Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longe 0) for the number of moles 98 n Solution, enter a zero OH Cion Ag mo mol Previous Give Up & View Solution Check Answer ONext SExit pyright 20 2015 Sapling Le 90 contact us help cy policy Cole Chegg.com Oth Resources Assignment Information Available From 10/15/2015 Due Date 0/22/20 Points Possible 00 Grade Category: Graded Description: Policies Homework You can check your answers You can view solutions when yo up on any question You can keep trying to answer you get it right or give up You lose 5% of the points availa in your question for each incorr answer. O Help With This Topic O Web Help & Videos O Technical Support and Bug R

Explanation / Answer

Solution containing 121.73 g of silver chlorate react with solution containing 10.19 g lithium hydroxide. How many gram of solid ppt will form.

Soltion-

Given-

Mass of AgClO3 = 121.73 g   molar mass = 191.31 g/mol

Mass of LiOH = 10.19 g    Molar mass 23.94g/mol

We know the equation to calculate # of mole

# mole = mass/molar mass

# mole of AgClO3 = 121.73/191.31 = 0.636 mol

# mole of LiOH = 10.19/23.94 = 0.425 mole

Let’s write the reaction

AgClO3 + LiOH AgOH (s) + LiClO3

Reaction is in stoichiometry

For the 121.73 g AgClO3 79.09 g of AgOH precipitate has to form

But only 10.19 g of LiOH is reacted for this AgOH precipitate = 53.13 g

Remaining precipitate of AgOH = 79.09 – 53.13

= 25.96 g.

Answer- 25.96 g. of solid precipitate was formed

How many gram of reactant in excess will remain after reaction?

Total reactant = 121.73g

Reactant used for 10.19 g of LiOH= 81.40 g

Excess reactant remaining = 121.73 g – 81.40 g = 40.33 g

Answer – 40.33 g of reactant remaining after reaction.

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