1 / 4

Basic Stoichiometry Problems

Basic Stoichiometry Problems. Please use these examples to solve today’s problems. Example #1. If 35.80g of aluminum chloride reacts with excess silver nitrate, how many grams of the ppt will be formed? 1 AlCl 3(aq) + 3AgNO 3(aq) Al(NO 3 ) 3(aq) + 3 AgCl (s)

muriel
Download Presentation

Basic Stoichiometry Problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Basic Stoichiometry Problems Please use these examples to solve today’s problems

  2. Example #1 If 35.80g of aluminum chloride reacts with excess silver nitrate, how many grams of the ppt will be formed? 1AlCl3(aq) + 3AgNO3(aq) Al(NO3)3(aq) + 3AgCl(s) 35.80g AlCl3 1mol AlCl33mol AgCl 143.32g AgCl 132.76g AlCl31mol AlCl3 1mol AgCl = 115.9g AgCl

  3. Example #2 15.0ml of a 3.0M magnesium chloride solution reacts with excess silver nitrate, how many grams of ppt will be formed? 1 MgCl2(aq) + 2 AgNO3(aq) Mg(NO3)2(aq) + 2AgCl(s) 15.0ml 1 Liter 3.0 mole MgCl22mol AgCl 143.32g AgCl 1000l 1 liter 1mol MgCl2 1mol AgCl = 12.9g AgCl

  4. Example #3 12.01g of ammonium carbonate reacts with excess ferrous bromide. How many grams of the ppt will be formed? 1(NH4)2CO3(aq) + FeBr2(aq) 2NH4Br(aq) + 1FeCO3 (s) 12.01g(NH4)2CO3 1mol (NH4)2CO31mol FeCO3 115.86g 96.11g 1mol (NH4)2CO3 1mol FeCO3 = 14.48g FeCO3

More Related