Chemical reactions

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Cl2 was the limiting reactant.

Therefore, Al was present in excess. But how much?

First find how much Al was required.

Then find how much Al is in excess.

How much of which reactant will remain when reaction is complete?

Slide 36

2 Al + 3 Cl2

products

0.200 mol

0.114 mol = LR

Calculating Excess Al

Excess Al = Al available - Al required

= 0.200 mol - 0.0760 mol

= 0.124 mol Al in excess

Slide 37

Determining the Formula of a Hydrocarbon by Combustion

CCR, page 138

Slide 38

Using Stoichiometry to Determine a Formula

Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O.

CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O

What is the empirical formula of CxHy?

Slide 39

Using Stoichiometry to Determine a Formula

First, recognize that all C in CO2 and all H in H2O is from CxHy.

CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O

0.379 g CO2

0.1035 g H2O

1 H2O molecule forms for each 2 H atoms in CxHy

1 CO2 molecule forms for each C atom in CxHy

Slide 40

Using Stoichiometry to Determine a Formula

First, recognize that all C in CO2 and all H in H2O is from CxHy.

1. Calculate amount of C in CO2

8.61 x 10-3 mol CO2 --> 8.61 x 10-3 mol C

2. Calculate amount of H in H2O

5.744 x 10-3 mol H2O -- >1.149 x 10-2 mol H

CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O

Slide 41

Using Stoichiometry to Determine a Formula

Now find ratio of mol H/mol C to find values of “x” and “y” in CxHy.

1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C

= 1.33 mol H / 1.00 mol C

= 4 mol H / 3 mol C

Empirical formula = C3H4

CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O