Chapter 7: Chemical Reactions Section 7.1 Describing Reactions Section 7.1 Describing Reactions

Chemical Reactions  aaw&feature=related aaw&feature=related  aaw&feature=related aaw&feature=related

Equations  Examples:  With words  carbon + oxygen > carbon dioxide  With formulas (a chemical equation)  C + O > CO 2  Examples:  With words  carbon + oxygen > carbon dioxide  With formulas (a chemical equation)  C + O > CO 2

The Law of Conservation of Mass  conservation of mass clip conservation of mass clip  conservation of mass clip conservation of mass clip

Balancing Equations  Look at this equation: N 2 H 4 + O 2 ---> N 2 + H 2 O Are the numbers of nitrogen atoms, hydrogen, and oxygen atoms the same on both sides? NO! We need to balance the equation. do this by changing the coefficients (the numbers that appear before the formulas). When there is no other, a coefficient of 1 is assumed.  Look at this equation: N 2 H 4 + O 2 ---> N 2 + H 2 O Are the numbers of nitrogen atoms, hydrogen, and oxygen atoms the same on both sides? NO! We need to balance the equation. do this by changing the coefficients (the numbers that appear before the formulas). When there is no other, a coefficient of 1 is assumed.

N 2 H 4 + O 2  N 2 + H 2 O  N=  H =  O =  N= H =H = O =O = N =N = H =H =  O = When you add a coefficient of 2, you get this balanced equation: N 2 H 4 + O 2  N 2 + 2H 2 O

More practice with balancing equations  Balance these three equations:  Cu + O > CuO  H 2 O > H 2 O + O 2  Mg + HCl > H 2 + MgCl 2  Balance these three equations:  Cu + O > CuO  H 2 O > H 2 O + O 2  Mg + HCl > H 2 + MgCl 2

Moles

A mole is a lot of things!  To comprehend the enormous size of Avogadro’s number, 6.02 x 10 23, here are some analogies.  Would a mole of rice grains fill our chemistry lab?  Would a mole of rice grains fit in our school?  1 mole of rice grams would cover all the land area of the whole world to a depth of 75 meters  To comprehend the enormous size of Avogadro’s number, 6.02 x 10 23, here are some analogies.  Would a mole of rice grains fill our chemistry lab?  Would a mole of rice grains fit in our school?  1 mole of rice grams would cover all the land area of the whole world to a depth of 75 meters

1 mole of rice has more grains than the number grains of all rice grown since the beginning of time!

1 mole of watermelon seeds:  Would be found in a watermelon slightly larger than the moon!

1 mole pennies divided equally between every person on earth:  Each person would receive 1 x pennies.  Personal spending at the rate of $1 million per day would use up each person’s wealth in just under 3000 years!  Life would not be comfortable. The surface of our planet would be buried in copper coins to a depth of about 420 m.  Each person would receive 1 x pennies.  Personal spending at the rate of $1 million per day would use up each person’s wealth in just under 3000 years!  Life would not be comfortable. The surface of our planet would be buried in copper coins to a depth of about 420 m.

Each human being has about 60 million body cells (6 x )  Assume that the earth’s population is 6 billion (6 x 10 9 ), the total number of living body cells on the earth at the present time is 3.6 x or a little more than half a mole!  Assume that the earth’s population is 6 billion (6 x 10 9 ), the total number of living body cells on the earth at the present time is 3.6 x or a little more than half a mole!

Molar Mass  CO 2 : carbon (12 g) + oxygen (2g x 16g) = 44g

Mole-Mass Conversions  Example:  You have 55 grams of CO 2. How many moles of CO 2 do you have?  Example:  You have 55 grams of CO 2. How many moles of CO 2 do you have?

More calculations….

Examples to work  How much oxygen is needed to make 144 grams of water?  two things are needed to work this problem ~ a balanced equation and molar masses.  first step is to find out how many moles of water we are trying to make. 144 g of H 2 O x 1 mol H 2 O = 8 mol H 2 O 18 g H 2 O  How much oxygen is needed to make 144 grams of water?  two things are needed to work this problem ~ a balanced equation and molar masses.  first step is to find out how many moles of water we are trying to make. 144 g of H 2 O x 1 mol H 2 O = 8 mol H 2 O 18 g H 2 O

Next step….  Next step - go to the balanced equation for a ratio of moles of oxygen to moles of water. 2H 2 + O > 2H 2 O 1 mol O 2 OR2 mol H 2 O 2 mol H 2 O1 mol O 2  Next step - go to the balanced equation for a ratio of moles of oxygen to moles of water. 2H 2 + O > 2H 2 O 1 mol O 2 OR2 mol H 2 O 2 mol H 2 O1 mol O 2

Next step….  convert moles of water you are trying to make and then to moles of oxygen needed to make that amount of water. 8 mol H 2 O x 1 mol O 2 = 4 mol O 2 2 mol H 2 O  convert moles of water you are trying to make and then to moles of oxygen needed to make that amount of water. 8 mol H 2 O x 1 mol O 2 = 4 mol O 2 2 mol H 2 O

Finally…..  Convert the moles of oxygen needed to the mass of oxygen needed. 4 mol O 2 x 32 grams = 128 grams 1 mol O 2 So, 128 grams of oxygen must be used in order to produce 144 grams of water.  Convert the moles of oxygen needed to the mass of oxygen needed. 4 mol O 2 x 32 grams = 128 grams 1 mol O 2 So, 128 grams of oxygen must be used in order to produce 144 grams of water.

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