Stoichiometry Mole-Mole Problems

11

Mr. ShieldsMr. Shields Regents Chemistry Regents Chemistry Unit 11 L03Unit 11 L03

22

Once we’ve balanced a chemical equationWhat 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 …OK. Let’s see …Hydrazine Hydrogen peroxideHydrazine Hydrogen peroxide

33

N2H4(g) + 2H2O2 (l) N2 (g) + 4H20 (l)

This equation provides the following information:

1. What’s reacting and what’s produced2. The states of matter involved in the reaction3. 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 thenWe can also state this equation in terms of moles.

So, Why is that?

44

2 molecules 1 molecule 2 molecules2 molecules 1 molecule 2 molecules

Which is: 2 x NWhich is: 2 x NAA 1 x N 1 x NA A 2 x N 2 x NAA..

Let’s look at the reaction between HLet’s look at the reaction between H22 and O and O22

It’s just a question of “scale”. It’s just a question of “scale”.

Whether it’s 2 molecules to 1 molecule or 2 moles to 1 mole

55

In or prior example 1 mole of hydrazine plus2 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 theLaboratory must we insure we have exactly1 mole of of hydrazine and 2 moles of hydrogenPeroxide present ?

No.No. Let’s see why… Let’s see why…

66

Whether we talk about1 molecule or 6.023 x 1023 molecules (1 mole) oreven 3.01 x 1023

It’s just a question of “scale”

We only need to insure the ratio’sremain the same

Remember… we said we can state chemicalequations in terms of molecules or in terms ofnumber of moles.

77

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 productWhy?

2N2H4(g) + 4H2O2 (l) 2N2 (g) + 8H20 (l)We must keep all the Mole ratio’s the same!

88

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)

So,So, 2 2NN22HH44(g) + (g) + 44HH22OO22 (l) (l) 22NN22 (g) + (g) + 88HH220 (l)0 (l)

1:1:11 And And 1:1:44 Lets look at Nitrogen first:If hydrazine is inc. to 2 mol N2H4 N2

Then 1 : 1 x = 2 2 : xFor water:

If hydrazine is inc. to 2 then 1 : 4 x = 8 2 : x

99

If I cut the number of moles of reactants in halfHow 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 productsAre also reduced by half. (what are the mole ratios involved?)

Let’s see how we would use this in solvingSome problems.

1010

What would you predict the number of moles ofReactant and products to be when 6 moles of Hydrazine reacts completely with H2O2?

6N2H4(g) + xH2O2 (l) xN2 (g) + xH20 (l)

1N2H4(g) + 2H2O2 (l) 1N2 (g) + 4H20 (l)

The ratio of N2H4 to H2O2 is 1:2 or 6:12The ratio of N2H4 to N2 is 1:1 or 6: 6The ratio of N2H4 to H2O is 1:4 or 6:24

OK… 1st you need to recall the balanced equation

Then we need to look at the change in mole ratios

1111

So our balanced equation changes from

N2H4(g) + 2H2O2 (l) N2 (g) + 4H20 (l)

6 N2H4(g) + 12 H2O2 (l) 6 N2 (g) + 24 H20 (l)

To

OK. That’s easy, right? Let’s try another one

1212

How many moles of N2H4 would be required toCompletely react 0.25 moles of hydrogenPeroxide? 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)

1313

N2H4(g) + 2H2O2 (l) N2 (g) + 4H20 (l)

x N2H4(g) + ¼ H2O2 (l) x N2 (g) + x H20 (l)

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: 0.125The ratio of H2O2 to H2O is 1:2 or 0.25: 0.5

Let’s look at the change in mole ratios of H2O2to each of the other reactant & Product molecules

So what is the final equation?..125 N2H4(g) + 0.25 H2O2 (l) .125 N2 (g) + 0.5 H20 (l)

1414

A Word Problem:

A chemist wants to produce 2.5 mol of water byreacting Hydrogen with Oxygen. How many molesOf each will he need?

2H2 + O2 2H2O (balanced eqn)

Ratio of H20 to H2 = 1:1Ratio of H20 to O2 = 2:1

2.5 mol of H2 and 1.25 mol of O2