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Mr. Shields
Regents Chemistry
Unit 11 L03
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Once we’ve balanced a chemical equation
What other information does if provide?
For example:
What information does the following give us?
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
OK. Let’s see …
Hydrazine
Hydrogen peroxide
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N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
This equation provides the following information:
1. What’s reacting and what’s produced
2. The states of matter involved in the reaction
3. How many molecules of each reactant
are needed for the reaction
4. How many molecules of product are produced
If we know how many molecules are involved then
We can also state this equation in terms of moles.
So, Why is that?
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Let’s look at the reaction between H2 and O2
2 molecules
Which is:
2 x NA
1 molecule
2 molecules
1 x NA
2 x NA.
It’s just a question of “scale”.
Whether it’s 2 molecules to 1 molecule
or 2 moles to 1 mole
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In or prior example 1 mole of hydrazine plus
2 moles of hydrogen peroxide yield
1 mole of Nitrogen and 4 moles of water.
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
If we react these chemicals together in the
Laboratory must we insure we have exactly
1 mole of of hydrazine and 2 moles of hydrogen
Peroxide present ?
No. Let’s see why…
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Remember… we said we can state chemical
equations in terms of molecules or in terms of
number of moles.
Whether we talk about
1 molecule or 6.023 x 1023 molecules (1 mole) or
even 3.01 x 1023
It’s just a question of “scale”
We only need to insure the ratio’s
remain the same
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So what would happen if I double the # of
Moles of reactants?
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
2N2H4(g) + 4H2O2 (l)  xN2 (g) + xH20 (l)
Right … I double the # of moles of product
Why?
2N2H4(g) + 4H2O2 (l)  2N2 (g) + 8H20 (l)
We must keep all the Mole ratio’s the same!
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In this simple problem let’s see how we would calculate the
new mole ratio’s. What is the ratio of N2H4 to N2 and N2H4
to H20 in the original balanced eqn.?
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
1:1 And 1:4
Lets look at Nitrogen first:
If hydrazine is inc. to 2 mol
N 2H 4
N2
Then 1 : 1
2 : x
x=2
For water:
If hydrazine is inc. to 2 then 1 : 4
2 : x
x=8
So, 2N2H4(g) + 4H2O2 (l)  2N2 (g) + 8H20 (l)
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If I cut the number of moles of reactants in half
How many moles of product will be produced?
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
½ N2H4(g) + 1H2O2 (l)  ½ N2 (g) + 2 H20 (l)
Right … the number of moles of the products
Are also reduced by half.
(what are the mole ratios involved?)
Let’s see how we would use this in solving
Some problems.
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What would you predict the number of moles of
Reactant and products to be when 6 moles of
Hydrazine reacts completely with H2O2?
6N2H4(g) + xH2O2 (l)  xN2 (g) + xH20 (l)
OK… 1st you need to recall the balanced equation
1N2H4(g) + 2H2O2 (l)  1N2 (g) + 4H20 (l)
Then we need to look at the change in mole ratios
The ratio of N2H4 to H2O2 is 1:2 or 6:12
The ratio of N2H4 to N2
is 1:1 or 6: 6
The ratio of N2H4 to H2O is 1:4 or 6:24
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So our balanced equation changes from
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
To
6 N2H4(g) + 12 H2O2 (l)  6 N2 (g) + 24 H20 (l)
OK. That’s easy, right? Let’s try another one
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How many moles of N2H4 would be required to
Completely react 0.25 moles of hydrogen
Peroxide? b) How man moles of product would be
formed?
Remember … first you need to know what the
Balanced equation is.
N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
x N2H4(g) + ¼ H2O2 (l)  x N2 (g) + x H20 (l)
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N2H4(g) + 2H2O2 (l)  N2 (g) + 4H20 (l)
x N2H4(g) + ¼ H2O2 (l)  x N2 (g) + x H20 (l)
Let’s look at the change in mole ratios of H2O2
to each of the other reactant & Product molecules
The ratio of H2O2 to N2H4 is 2:1 or 0.25 : 0.125
The ratio of H2O2 to N2
is 2:1 or 0.25: 0.125
The ratio of H2O2 to H2O is 1:2 or 0.25: 0.5
So what is the final equation?
.125 N2H4(g) + 0.25 H2O2 (l)  .125 N2 (g) + 0.5 H20 (l)
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A Word Problem:
A chemist wants to produce 2.5 mol of water by
reacting Hydrogen with Oxygen. How many moles
Of each will he need?
2H2 + O2  2H2O (balanced eqn)
Ratio of H20 to H2 = 1:1
Ratio of H20 to O2 = 2:1
2.5 mol of H2 and 1.25 mol of O2
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