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Chapter 7 Chemical reactions. Signs of a chemical reaction color change (ex. bleached hair, steel rusting) solid forms see bubbles heat produced (temp.

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Presentation on theme: "Chapter 7 Chemical reactions. Signs of a chemical reaction color change (ex. bleached hair, steel rusting) solid forms see bubbles heat produced (temp."— Presentation transcript:

1 Chapter 7 Chemical reactions

2 Signs of a chemical reaction color change (ex. bleached hair, steel rusting) solid forms see bubbles heat produced (temp change)

3 Chemical equations CH 4 + O 2  CO 2 + H 2 O ex. of a chemical reaction chemical reactions are represented by chemical equations reactants are to the left of the arrow and products to the right of the arrow reactants  products

4 Chemical equations CH 4 + O 2  CO 2 + H 2 O same atoms to the left and right of arrow, only the bonding has changed In a chemical reaction atoms are neither created nor destroyed same number of each type of atom to the left and right of a chemical equation. Making sure this happens is called balancing the equation

5 Balanced equation CH 4 + O 2  CO 2 + H 2 O [unbalanced] CH 4 + 2O 2  CO 2 + 2H 2 O [balanced] besides specifying compounds involved, we also specify the state of matter for each compound Symbolstate (s)solid (l)liquid (g)gas (aq.)dissolved in water

6 states of matter CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l)

7 balancing equations You may NEVER change subscripts or add atoms. You can add coefficients in front of compounds only! Step 1: Put a 1 as the coefficient for ALL compounds in a chemical equation

8 balancing equations C 2 H 5 OH + O 2  CO 2 + H 2 O Step 1 Put a 1 as the coefficient to ALL cmpds in a chemical equation. So the above reaction becomes: 1 C 2 H 5 OH + 1 O 2  1CO 2 + 1 H 2 O Next, fill out a RAP table

9 RAP Table R= reactants A= atoms P= products

10 RAP example For the equation: 1 C 2 H 5 OH + 1 O 2  1CO 2 + 1 H 2 O RAP 2C1 6H2 3O3

11 Balance Balance the equation by changing the coefficients and then adjusting #s in the RAP table. Your goal is to have the same # of each atom on the R and P columns!

12 For the equation: 1 C 2 H 5 OH + 1 O 2  1CO 2 + 1 H 2 O First start by changing the CO 2 coefficient to 2. This will change the # of carbons on the right to 2 AND the # OF OXYGENS ON THE RIGHT TO 5. RAP 2C1 6H2 3O3

13 1 C 2 H 5 OH + 1 O 2  2CO 2 + 1 H 2 O Now let’s change the water coefficient from 1 to 3 to balance the hydrogens. RAP 2C2 6H2 3O5

14 1 C 2 H 5 OH + 1 O 2  2CO 2 + 3 H 2 O Lastly let’s change the coefficient of O 2 to 3. RAP 2C2 6H6 3O7

15 Balanced!!! 1 C 2 H 5 OH + 3 O 2  2CO 2 + 3 H 2 O Notice both the R AND P columns match for ALL atoms. This means the equation is BALANCED!!! RAP 2C2 6H6 7O7

16 Tricks to using this method 1. polyatomic ion trick: –If you see a polyatomic ion on BOTH the reactant AND product side of the equation, do NOT split the polyatomic ion up into atoms in the RAP table. ex: Cu + AgNO 3  Cu(NO 3 ) 2 + Ag Nitrate shows up on the left AND right of the arrow, so be sure to keep nitrate together in the RAP table. Do NOT split nitrate up into N and O.

17 Tricks to using this method 2. ½ trick –Sometimes it is helpful to use ½ numbers instead of whole numbers while balancing equations. Be sure to get rid of ½ numbers at the end by multiplying ALL coefficients by 2. Remember that a balanced equation MUST have the lowest whole number ratio to be correct! ex. while balancing: CH 3 OH + O 2  CO 2 + H 2 O it is helpful to change the water coefficient to 2 to balance the Hs and then if you put a 1.5 in front of O 2 the equation is technically balanced and looks like this: 1 CH 3 OH + 1.5 O 2  1 CO 2 + 2 H 2 O Now multiply ALL coefficients by 2 to get rid of the half. Which leaves you with the following balanced equation: 2 CH 3 OH + 3 O 2  2 CO 2 + 4 H 2 O (correct answer)

18 Tricks to using this method 3. water trick: –If you see H 2 O on one side of an equation and OH on the other it is sometimes helpful to change H 2 O into H(OH). ex: HCl + Ca(OH) 2  H 2 O + CaCl 2 by changing H 2 O into H(OH), you can see that there is hydroxide on both sides of the equation and use this to balance.

19 Summary The tricks that I’ve given can be helpful, but many times people like using a trial/error approach to balancing. As long as you have the lowest whole # ratio of coefficients, I don’t care if you use trial/error or RAP method.

20 You must also be able to take a word equation and change it to compound formulas and then balance. An example of a word problem is: Mercury (II) nitrate solution reacts with potassium iodide solution to give a mercury (II) iodide precipitate and potassium nitrate solution. The word “reacts” and “and” indicates a + sign. “to give” or “yields” always is the arrow. So this equation looks like the following (unbalanced): Hg(NO 3 ) 2 + KI  HgI 2 + KNO 3 The word “precipitate” indicates a solid (s) material. The word solution indicates an aqueous material (aq.). So, the real equation unbalanced looks like: Hg(NO 3 ) 2 + KI (aq)  HgI 2(s) + KNO 3(aq)

21 word problems Other important words in word problems: –“metal” means that element all by itself –oxygen gas= O 2(g) –iodine = I 2 NOT I –fluorine = F 2 NOT F –nitrogen = N 2 NOT N –All diatomic elements have the same name as the element that makes up the diatomic molecule!!

22 Other word problem examples magnesium metal reacts with chlorine to produce magnesium chloride. ???? zinc reacts with hydrochloric acid to produce hydrogen gas and zinc chloride. ????


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