Limiting Reactants

- usually 1 reactant is used up before the other
- this reactant is called the limiting reactant (or reagent) ----> it stops or limits the reaction.
- the limiting reactant determines the amount of products produced.
- assume one reactant is the limiting reactant, and determine what quantity of the other is needed.

example

what is the limiting reactant when 25 g of p₄ reacts with 64.6 g of Cl₂ forming phosphorus trichloride?
P₄ + 6 Cl₂ ---> 4PCl₃
P₄ = 123.6 g/mol
Cl_{2 = 71 g/mol}


Since you need more than the given amount of Cl₂ to fully react the P₄, Cl₂ is the limiting reagant.


Determine the theoretical yield of the above reaction.
PCl<sub>3 = 137.4 g/ mol

when we figure out the theoretical yield, we have to calculate the amount of the product that will be produced from the limiting reagent, because that is how much is being used before the reaction is over.</sub>







83.3 g of PCl3 will be produced. (we used the limiting reagent in the equation because that is the amount that will be used in the reaction.


example.
in the formation of water, 25g of oxygen reacts with 10g of hydrogen. determine the limiting reactant, theoretical yield and amount of excess reactants.

2O2 + 2H2 -----> 2 H2O

H2 = 2 g/ mol
O2 = 32 g/mol

For 10g of hydrogen, we need 80 g of oxygen, so
O2 is the limiting reagent.



H2O = 18 g / mol
THEORETICAL YIELD
28.1 g of water will be produced.


EXCESS - we plug in the calculation so we know exactly how much hydrogen we need, then we subtract this amount from the amount that is given to know how much excess there is.

10g - 3.13g = 6.87 g will be left.

***** Notice that the amounts you need (25 g O2 + 3.13g H2 = 28.1 which is the theoretical yield!!!)



here's a worksheet

and heres a video for further help:

Stoichiometry Lab

In class we tested if stoichiometry could predict precipitate. The materials we used were:
-a lab stand
-a funnel
-a clamp
-filter paper
-beakers
-squirt bottle
To test stoichiometry we used 3 grams of Copper(II) Sulfate and 2 grams of Strontium Nitrate. We dissolve those two in seperate beakers and mixed them together. After stirring for about 7-10 minutes we carefully poured the mixture into the filter. We diluted and continued to dissolve the leftover precipitate. When the mixture was all dissolved we put the filter paper in the dryer to leave over night

Mole to Mass and Other Conversions

Examples
1. Lead(II) Nitrate reacts with 5.0g of potassium iodide. How many grams of Lead(IV) nitrate are required for a complete reaction?

Step 1. Find balanced equation
Pb(NO3)4 + 4 KI-> PbI4 + 4 KNO3

Step 2. Convert and multiply what you need over what you're given
5.0g x 1mol/166g x 1Pb(NO3)4/4KI x 455.2g/1mol Pb(NO3)4 = 3.5g


2. How many grams of O2 are produced from the decomposition of 3.0g of potassium chlorate?

Step 1. Find balanced equation
2 KClO3 -> 2 KCl + 3 O2

Step 2. Convert and multiply what you need over what you're given
3.0g x 1mol/122.6g x 3mol of O2/2mol of KClO3 x 32g/1mol of O2 = 1.2g


3. If a 100mL solution of 2.0M H2SO4 is neutralized by sodium hydroxide. What mass water is produced?

Step 1. Find balanced equation
H2SO4 + 2 NaOH -> 2 HOH + Na2SO4

Step 2. Convert(in this case use concentration formula to find the number of mol) and multiply what you need over what you're given
0.100L x 2.0mol/1L = 0.200mol H2SO4 x 2mol HOH/1mol H2SO4 x 18.0g/1mol = 7.2g

*Remember for next class' lab*
-Theoretical yield of a reaction is the quantity of products expected(calculation)
-The amount produced in an experiment is the actual yield
-The percent yield is : %yield = actual/theoretical x 100

More Stoichiometry

Remember last class when Mr. Doktor introduced us to the wonderful work of stoichiometry?(If you don't we wrote a blog about that one too! haha) Well today we learnt even more!! In adition to moles to mass conversions, we learnt how to solve mass to mole conversions. We also learnt how to solve problems involving volume at STP(22.4L). To help you remember info about volume at STP the chart below could be very handy. You can use this chart to help you by seeing which order to convert your elements and by how much.

Here's a quick example..

What mass of water vapour is produced when 3.5L of Hydrogen is burnt(with oxygen) at STP?

Click here for a few pratice questions!!

And here's a little video of a guy with a cool accent to try to explain it for ya!!

Stoichiometry

so beginning today, we will begin to put together all the things we have learned in the last couple months so that they work together. anything from mole conversions to balancing equations will be involved and it sounds like pretty interesting stuff!

mole to mole calculations

• coefficients in balanced equations respresent the number of moles in each part
• they can also be used as conversion factors

example:

if .15 mol of methane reacts with oxygen, how many moles of water are produced?

first we find the balanced equation:

CH4 + 2O2 -> CO2 + 2H2O

then we find out how many moles of water are needed for the equation by using the formula:

(what you're given) X (what you need)/(what you know)

so... (.15 moles of methane) X (2 moles of water)/(1 mole of methane) = .30 moles of H2O

so by using the quantity we are given, we can find out any other of the factors in the equation

we can also take it one step farther by multiplying the mole ratio we found by the molar mass of the chemical we are trying to find the mass of

for example we can take the .30 moles of water and multiply that by its molar mass (18.02 g/mol) and we find that 5.4 g of water has been used in the equation.

so thats all the notes for today!

Candle Wax Lab

In class we tested how much energy a small candle can release!

The material needed are:
1) candle
2) matches
3) weigh scale
4) lab stand
5) wire mesh
6) thermometer
7) water

Our observations were:
1) volume of water used: 100mL
2) inital temperature of water: 26 degrees C
3) inital mass of candle: 10.1 g
4) final temperature : 42 degrees C
5) final mass of candle: 9.80 g

In this experiment 0.000880 mole of paraffin wax was reacted.
6.704 kJ was the amount of energy gained by the water. The molar heat of combustion of Paraffin wax is 7.6 x 10 to the power of 3 kJ/mol

Calorimetry

To know the amount of energy released three things must be known.
1) Temperature change (▲T) - measured with a thermometer
2) Amount of water (kg) ..... 1 g = 1 mL.... 1kg = 1000mL - measured with a scale
3) Specific heat capacity - how much heat a given substance can hold
water can hold 4.19 kj/kg °C

Equation
▲H= mC▲T
▲H= enthalpy change (kj)
m= mass of water (kj)
C = specific heat capacity (kg/kj °C)
▲T= change in temperature (°C)

Molar Enthalpy - Change in heat for each mole reacted

If you burn 0.315 moles of hexane (C₆H₁₄) in a bomb calorimeter containing 5.65 kg of water, what’s the molar heat of combustion of hexane is the water temperature rises 55.4⁰ C?

▲H = mC▲T

▲H = (5.65 kg)(4.19 kj/kg⁰C)(55.4⁰ C)

▲H = 1311.5 kJ

= 1312 kJ

If you burn 22.0 grams of propane (C₃H₈) in a bomb containing 3.25 kg of water, what’s the molar heat of combustion of propane if the water temperature rises 29.5⁰ C?

▲H = mC▲T

▲H = (3.25 kg)(4.19 kj/kg⁰C)(88.5⁰ C)

▲H = 1205.15 kj

▲H = 1205 kj

here is a quick worksheet to test your skills!

and here is a quick video that shows an experiment involving enthalphy. It is an exothermic reaction



and here is a how to video if youre still confused

Heat in Chemical Reactions

Here are a few things to remember...

• reactions that release heat are exothermic
• reactions that absorb heat are endothermic
• heat IS a form of energy
• all chemicals have stored energy, also known as enthalpy
• enthalpy is given the symbol 'H'
• ΔH is the change in enthalpy

Enthalpy Diagrams

Exothermic

• high to low enthalpy
• ΔH is negative
• heat is released

Endothermic

• low to high enthalpy
• ΔH is positive
• heat is absorbed

wanna learn more about enthalpy change? click here....... you know you waaaaaannaa

--
this is totally off topic but i think Mr. Doktor will find this pretty cool. well at least i think its pretty awesome..its a tv show on discovery or FSN called sport science. Hope you enjoy :)

Balancing Chemical Equations!

There are 6 types of Chemical Reactions!

1 SYNTHESIS: A + B = AB
Al + Cl = AlCl3 The balanced equation is: 2Al + 6Cl = 2AlCl3

2 DECOMPOSITION: AB = A + B
AlBr3 = Al + Br2 The balanced equation is: 2AlBr3 = 2Al + 3Br2

3 SINGLE REPLACEMENT: A + BC= B + AC
Ag(No3) + Zn = Ag + Zn(No3)2 The balanced equation is: 2Ag(No3) + Zn = 2Ag + Zn(No3)2

4 DOUBLE REPLACEMENT: AB + CD = AD + CB
MgCl2 + K2So4 = MgSo4 + KCl The balanced equation is: MgCl2 + K2So4 = 2KCl

5 NEUTRALIZATION: Acid + Base = Water + Ionic Salt
Ba(OH)2 + HCl = BaCl2 + H2O The balanced equation is: Ba(OH)2 + 2HCl = BaCl2 + 2H2O

6 COMBUSTION: Reaction with Oxygen
Mg + O2 = MgO The balanced equation is: 2Mg + O2 = 2MgO

For more practice here is a link to a worksheet!
"http://www.bishops.k12.nf.ca/science/1206/chem/reactiontypes.htm"

Balanced Equations

K so we went over all the stuff about balanced equations that we were talking about before the break. Not many notes for this class, just a lot of examples so ill use some of the notes from last year to supplememnt.

steps to balancing:

• first we must make sure the equation is correct

Mg + O2 - MgO

• balance the equation so there are the same amount of atoms on each side

2Mg + O2 - 2MgO

here are some acids that me are going to have to remember for the acid neutralization reactions:

• HCl: hydrochlroic acid
• HNO3: nitric acid
• H2SO4: sulphuric acid
• H3PO4: phosphoric acid
• CH3COOH: acetic acid

k so this video is pretty long but it gets the point across so s'all good :P

thats pretty much all we did today. ttfn!