Plz. help another set of problem for my chemistry..?

Question

In the chemical reaction:

C3H8+5O2----->3CO2+4H2O

Suppose that 1.5 moles of C3H8 will be used.

1. How many grams is 1.5 moles C3H8(propane)?

2.How many moles of H2O is produced?

3. How many grams CO2 is formed?

4. How many molecules of H2O are formed?

5. If 6.75 moles of O2 and 3.25 moles C3H8 were combined. how many grams of H2O will be produced? Which is the limiting reagent?

6. If 100 grams each of O2 and C3H8 were combined . how many moles CO2 will be produced? Which is the excess reagent? How much excess reagent will be left unreacted?

numbers are the subscripts ^^ tnx!!

Answer 1

Ok, you need to understand that this limiting reagent problem involved stoichiometric ratios.

These ratios can be found when you write a correctly balanced equation.

This means the ratio of C3H8 to H20 is a 1: 4 ratio.

This also means that C3H8 to CO2 is a 1 : 3 ratio.

In limiting reagent problems the goal is to find the reacting compound which produces the LEAST amount of product. That would be the limiting reagent.

In order to find 'excess' reagent, you consider the reactant which is NOT fully consumed. This will be the other reactant which is NOT the limiting reagent. You take the initial amount of this, subtract the amount it took to react, and get the excess.

Also understand that on the periodic table, the molar mass listed for each element pertains to the amount found in 1 mole of this substance.

A mole is defined as 6.022 x 10^23 particles, atoms, molecules, etc. It is also called Avogadros number.

First:

1) moles to grams of C3H8

Molar mass of C3H8: 44.097g

You have 1.5 moles x (44.097g C3H8/1 mole C3H8)

=66.15g of C3H8

2) ? moles of H2O produced

**use the ratio

1.5 moles C3H8 x (4 moles H20/1 mole C3H8)

= 6 moles H2O

***This means that for every 1 mole, 4 moles of H2O are produced. Therefore for 1.5 moles, 6 moles are produced.

3) ? grams of CO2 are produced

*Use ratio, then convert to grams by using molar mass

1.5 moles C3H8 x (3 molesCO2/1 mole C3H8)

=4.5 moles CO2

Then you take this and multiply by the molar mass:

4.5 moles CO2 x (44.01g CO2/1 mole CO2)

=198.041

4) ? molecules H20

* do ratio and multiply by avogadros number

You already found the moles of H2O above which was:

6 moles H2O x (6.022x 10^23 molecules / 1 mole H2O)

=3.61 x 10^24 molecules

5) Carry out calculations separately.

a) 6.75 moles O2 x (4 moles H2O/5 moles O2) = 5.40 moles H2O

5.40 moles H2O x (18.015g H2O/1 mole H2O)

= 97.28 g H2O

b) 3.25 moles C3H8 x (4 mole H2O/1 mol C3H8) =14 moles H2O

14 moles H2O x (18.015g H2O/1 mol C3H8)

= 252.21g H2O

**The one that produces the least amount of grams of water is the limiting reagent.

6) convert grams to moles, mole to mole ratio, then back to grams.

100 g O2 x (1 mole O2/15.994)

=6.25 mole O2

6.25 mol O2 x (3 mol CO2/5mol O2)=

3.75 mol CO2

100 g C3H8 x (1 mol C3H8/ 44.097g C3H8)=

2.27 mol CO2

Therefore C3H8 is the limiting reagent and the excess of Oxygen is:

(there are a couple ways to do this based on ratios but this one is straight forward)

2.27 mol CO2 x (5 mol O2/3 mol CO2 x 15.999/1 mol O2)

=60.53g O2

Subtract this from 100g you started with and you have: 39.47g O2

You must work hard to understand why the ratios can help lead you from grams to moles to grams to even moles to molecules in order to fully do well in higher chemistry. Understand why a limiting reagent would produce an excess of the other element reacting. Also, write things out the way I did do that you can 'cancel' out the one on bottom. If you write out the way I did it, but with the numbers below the / sign on bottom, you'll see writing units will make sense. It's hard to explain but in your chem book pay attention to the 'conversion analysis' tips.

Ex:

20 g x 1 mol

.  .  .  . ______          =    .5 mol

.   .   .  .  40 g

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