# Reaction Stoichiometry: Chemical Equations, cont.

Notice that a chemical equation gives both qualitative information (what things react to give what products) and quantitative information (how much stuff is produced if a particular amount of stuff reacts). This gives rise to various practical applications.

• Example: How many grams of carbon dioxide are produced during the complete combustion of 475.5 g of natural gas (methane, CH4)

First need to find moles of CH4:

MW = 12.01 + 4 (1.008) = 16.04

Moles of CH4 = 475.5 g / 16.04 g/mol = 29.64 mole.

Next need balanced equation:

CH4 + 2 O2 CO2 + 2 H2O

Note from the equation we have a 1:1 ratio of moles methane to mole carbon dioxide

So we have 29.64 moles CO2

& MW of CO2 = 12.01 + 2 (16.00) = 44.01

and (29.64 mol)(44.01 g/mol) = 1.304 kg

Let's look at a slightly more complicated reaction.

• Example: How many moles of oxygen will be needed for the complete combustion of 275.5 g of butane (C4H10).

First we need the MW of butane: MW = 4 (12.01) + 10 (1.008) = 58.12 g/mol

and the number of moles of butane = (275.5 g) / (58.12 g/mol) = 4.740 mole

Then from the balanced equation:

2 C4H10 + 13 O2 8 CO2 + 10 H2O

we can find the number of moles of oxygen required

O2 = (4.740 moles butane)(13 moles O2 / 2 moles butane) = 30.81 moles

# Limiting Problems

Asking question of what is the maximum amount of something which can be produced from a given mixture of stuff. This is a fairly straight-forward sort of problem in the day-to-day world, but seems to cause a great deal of difficulty for lots of folks in chemistry. Let's start by looking at a non-chemical problem:

Consider you have to make a bunch of sandwichs for a party. The equation for the sandwichs (in slices) is:

2 Bread + 1 Cheese + 1 Meat 1 Sandwich

You have a 32 oz loaf of bread, 22 oz of sliced cheese and 32 oz of sliced meat. If bread slices weigh 1/2 oz, cheese slices 3/4 oz and meat slices 1 oz, how many sandwichs can you make?

Look at how many sandwichs can be made from each ingredient:

• If cheese limits: (22 oz cheese)(1 slice cheese/3/4 oz cheese)(1 sandwich/slice cheese) = 29.3 sandwiches
• If meat limits: (32 oz meat)(1 slice meat/oz meat)(1 sandwich/slice meat) = 32 sandwiches

Cheese limits and we can make 29 sandwichs.

• Example: Consider the reaction:

Fe3O4 + 4 C 3 Fe + 4 CO

What is the maximum mass of Fe which could be made from 115.0 g Fe3O4 and 24.00 g C?

The trick here is to find the maximum amount of iron which could be made from each reactant.

The lesser amount will then be the max possible:

C: (3 mol Fe/ 4 mol C)(24.00 g C/ 12.01 g C/mol C) = 1.499 mole

Fe3O4: (3 mol Fe/ mol Fe3O4)(115.0 g Fe3O4/231.6 g Fe3O4/mol Fe3O4) = 1.490 mole

therefore Fe3O4 limits, can only make 1.490 moles Fe.

Grams = (1.490 moles Fe)(55.85 g Fe/mol Fe) = 83.217 g Fe = 83.22 g Fe

• Example: Consider the reaction of zinc metal with acid:
Zn + 2H+ Zn2+ + H2(g)

What is the maximum amount (moles) of hydrogen gas which may be produced by reacting 0.50 g of Zinc with 0.800 mole hydrogen ion? Show work!

For Zn limiting: (1 mole H2/1 mole Zn)(0.50 g Zn) / (65.39 g Zn/mol) = 7.646 x 10-3 mol H2

For H+ limiting: (1 mole H2/2 mole H+)(0.800 mole H+) = 0.400 mol H2

Much less with Zn therefore, 7.6 x 10-3 mol H2

## Percent Yield

Another frequent question arising in chemical processes is the percent yield. This deals with the question of how effective was a given process in producing a product. Its an important consideration because chemical reactions rarely go completley to products. The maximum possible yield for a reaction is known as the Theoretical Yield.

• Example: Looking at the Zn limiting case above, the amount of hydrogen generated would be the theoretical yield = 7.6 x 10-3 mol. Now if the actual yield turned out for a particular experiment to be 6.75 x 10-3 mol, the percent yield would be calculated to be:

(6.75 x 10-3 mol)(100 %) / (7.6 x 10-3 mol) = 88.8% = 89%

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