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Friday, December 04, 2009 I am leaving last weeks web up for now so you can try to figure out these problems For Monday: Lab Reports All entries on the Calculation table must be supported by labled calculations! In addition do p. 321 # 25. and 26 Tuesday, December 01, 2009 I decided to start fresh. Here is how you do the homework sheet. 1. Write the balanced equation 2. Below the equation put the recipe that relates 3. Above the equation put the problem Example:
1. How many moles of oxygen may be produced from the decomposition of 3.6 moles of potassium chlorate?
problem 3.6 mol X =5.4 mol Want = Given
equation 2KClO3 --> 2KCl + 3O2 or Mol O2 = 3.6 moles KClO3 x 3moles O2 recipe 2 mol 3mol 1 2 mol KClO3 Do the first 5 problems, we will go over the Lab Report Tomorrow Wednesday, December 02, 2009 For tonight solve the following problems 1. In the lab you reacted sodium bicarbonate with hydrochloric acid. If you used 2.00g of sodium bicarbonate what mass of salt should you produce? How many moles of salt did you produce? What if you used 4.0 g of sodium bicarbonate how many grams of salt would you produce. Now try to answer questions 1-8 on your Lab. 2. When you breathe you take in oxygen to burn glucose (C6H12O6) and you produce carbon dioxide and water. If you burned 2000g of glucose what mass of products would be formed? 3. What mass of iron would be fromed from the reaction of 0.25 moles of aluminum with 10.0 g of iron III sulfate? Thursday, December 03, 2009 For tonight work on your lab report that is due on Monday and the above problems. I will work #3 below. 10.0g .25mol Fe2(SO4)3 + 2Al --> Al2(SO4)3 + 2Fe 399.5g 2moles You need to find which reactant is in excess or which one is the limiter. The real recipe is 399.5 g of iron III sulfate combines with 2 moles of Al. So if you had 10.0 g of iron III sulfate how many moles of Al would be consumed? Want= Given Moles Al = 10.0 g iron III sulfate x 1 mol iron III sulfate x 2 moles Al 1 399.5g 1 mol iron III sulfate =.05moles of Al Since they gave you .25moles of Al and you only need .05moles Al there is an excess of .20moles Al. The .25 can not be used in the stochiometry because it does not conform to the real recipe. 10.0g .05 moles X_________g Fe2(SO4)3 + 2Al --> Al2(SO4)3 + 2Fe 399.5g 2moles 2(55.8g) 111.6g There are two ways to solve for X 1. 10.0g x 111.6g / 399.5g = 2.79g 2. .05moles x 111.6g/2moles =2.79g