- WELCOME TO PART ONE OF THE VIDEO TUTORIAL SERIES
ON CHEMICAL REACTIONS.
IN THIS FIRST TUTORIAL,
WE WILL LOOK AT BALANCE IN CHEMICAL REACTIONS.
LET'S GET STARTED.
SO CHEMICAL REACTIONS OBEY THE LAW OF CONSERVATION OF MASS,
THAT MEANS THAT ATOMS ARE NEITHER CREATED NOR DESTROYED.
ALL RIGHT.
SO WHAT'S HAPPENING RIGHT,
IS THAT THE CHEMICAL BONDS ARE BEING BROKEN AND FORMED.
OKAY. SO REALLY A CHEMICAL REACTION IS SIMPLY
ABOUT REARRANGING HOW THE ATOMS ARE CONNECTED
THROUGH THE BONDS, ALL RIGHT.
SO LET'S LOOK AT AN EXAMPLE.
SO HERE, WE HAVE METHANE REACTING WITH OXYGEN
TO PRODUCE CARBON DIOXIDE AND WATER.
IF WE LOOK AT THIS FROM A VERY PHYSICAL PERSPECTIVE,
WE COULD LOOK AT THE MOLAR MASS OF METHANE SO--
AND SINCE THERE'S ONE MOLE REACTING
WE WOULD PUT 16.05 GRAMS,
WE WOULD HAVE TWO MOLES OF OXYGEN,
SO THAT WOULD CORRESPOND TO 64 GRAMS OF OXYGEN
AND THEN WE CAN LOOK AT OUR ARROW LIKE WE WOULD,
AN EQUAL SIGN OF MATHEMATICS AND IF WE LOOK AT,
THERE'D BE ONE MOLE OF CARBON DIOXIDE 44.01 GRAMS,
AND THEN LAST BUT NOT LEAST TWO MOLES OF WATER,
33.04 GRAMS, AND THEN WE DO PRODUCE HEAT AND WE'LL
TALK ABOUT IN A LATER TUTORIAL LOOKING AT THE ENERGY CHANGES
THAT OCCUR WITH CHEMICAL REACTIONS.
SO IF WE LOOK HERE WE SEE THAT ALL TOGETHER,
WE STARTED WITH 80.05 GRAMS OF MATERIAL
AND WHEN WE'RE ALL DONE,
WE HAVE 80.05 GRAMS.
SO WE SAY THE MASS IS CONSERVED.
WE CAN ALSO LOOK AT IT FROM THE POINT OF THE ATOMS,
ON THE LEFT WE SEE WE HAVE ONE CARBON,
FOUR HYDROGENS AND FOUR OXYGENS.
AND ON THE RIGHT WE HAVE ONE CARBON,
FOUR HYDROGENS AND FOUR OXYGENS.
SO WE HAVE ALSO MAINTAINED THE SAME NUMBER OF ATOMS
SO THAT'S WHAT WE TALK ABOUT THAT ATOMS
ARE NOT CREATED NOR DESTROYED.
ALL RIGHTY.
NOW, LET'S GO INTO THE ACTUAL TECHNIQUE
OF BALANCING CHEMICAL REACTIONS.
SO LINGUISTICALLY THE COEFFICIENT IS THE NUMBER
PLACED IN FRONT OF THE CHEMICAL REACTION,
RIGHT, SO THE COEFFICIENTS ARE HERE, NOTICE,
THAT WE NEVER SHOW THE ONES,
THE ONES ARE UNDERSTOOD.
ALL RIGHT.
NOW WE HAVE THE OPTION TO PUT IN SUBSCRIPTS
TO INDICATE THE PHYSICAL STATE
OF THE REACTANTS IN PRODUCTS AND YOU'VE SEEN THESE BEFORE,
RIGHT, S FOR SOLID, G FOR GAS, L FOR LIQUID AND AQ
WHEN WE'VE DISSOLVED THE SUBSTANCE IN WATER.
SO THE RULES FOR OUR CORRECT
CHEMICAL EQUATION ARE VERY STRAIGHTFORWARD.
IT HAS TO BE CONSISTENT WITH THE EXPERIMENTAL FACTS,
WE'RE ONLY GOING TO LIST REACTANTS AND PRODUCTS
IF THEY'RE ACTUALLY INVOLVED IN THE REACTION
AND WE ALWAYS HAVE TO USE THE CHEMICAL FORMULA,
CORRECT CHEMICAL FORMULA.
NOW, REMEMBER THAT THE CHEMICAL FORMULAS
COME FROM THE VALENCE ELECTRONS AND THE OCTET RULE.
SO WE DON'T HAVE THE FREEDOM TO CHANGE THAT,
THAT'S OUT OF OUR CONTROL.
AND SO TO BALANCE, IT'S FOCUSING ON THE COEFFICIENT'S,
THE COEFFICIENT'S GIVE US THE RATIOS
AND WE MUST OBEY THE CONSERVATION OF MASS.
SO NOW THAT WE HAVE AN OVERVIEW,
LET'S PRACTICE BALANCING SOME REACTIONS. OKAY.
SO BASICALLY THE FIRST THING WE WANT TO DO IS THIS--
IS LOOK BACK AND ASSESS THE EQUATION,
LOOKING AT THE NUMBER OF ATOMS
OF EACH ELEMENT ON THE REACTANT IN THE PRODUCT SIDE,
RIGHT IF THE TWO NUMBERS DO NOT MATCH THEN IT'S NOT BALANCE,
SO WE GOT TO GET TO WORK, ALL RIGHT.
SO WE'RE GOING TO BALANCE ONE ELEMENT AT A TIME,
ALL RIGHT, BY ADDING COEFFICIENTS, RIGHT.
WE NEVER CHANGE THE CHEMICAL FORMULA, RIGHT,
THAT COMES FROM VALENCE ELECTRONS AND THE OCTET RULE.
SO WE'RE ALL WISE AND POWERFUL
BUT WE DON'T GET TO MESS WITH THE OCTET RULE.
A COEFFICIENT APPLIES TO--
RIGHT TO EVERY ELEMENT IN THE FORMULA,
SO IT'S A BLENDING OF ENGLISH AND MATH,
IT'S A GOOD IDEA TO SAVE SINGLE ELEMENTS--
SINGLE ELEMENT PRODUCTS AND REACTANTS FOR LAST
BECAUSE YOU'RE ONLY INFLUENCING ONE ELEMENT--
ONE ELEMENT AT A TIME.
AND THEN REMEMBER THAT THE COEFFICIENT'S ALWAYS
NEED TO BE THE SMALLEST SET OF WHOLE NUMBERS.
WHEN YOU'RE ALL DONE, STEP BACK AND LOOK,
MAKE SURE YOUR COEFFICIENTS
CANNOT BE DIVIDED BY A COMMON FACTOR.
ALL RIGHTY.
NOW LET'S PRACTICE TOGETHER. ALL RIGHT.
SO HERE WE HAVE ALUMINUM REACTING
WITH OXYGEN TO PRODUCE ALUMINUM OXIDE,
SO SINCE OXYGEN AND ALUMINUM ARE BOTH BY THEMSELVES,
WE'LL FOCUS ON WHERE THEY'RE MIX TOGETHER AND WE COULD GO-- OOH,
THERE'S TWO OF THOSE, WE'LL PUT A TWO THERE.
AND LET'S--
SO LET'S JUST DO THAT, ALL RIGHT SO WE PUT A TWO
IN FRONT OF THE ALUMINUM AND THEN WE'RE LIKE,
HMM, OXYGEN, THERE'S THREE BUT IT'S DIATOMIC.
SO WE'LL HAVE TO PUT A THREE OVER TWO, RIGHT,
BECAUSE 3/2 X 2/1, THE TWOS
WILL CANCEL AND WILL END UP WITH THE THREE OXYGENS.
BUT NOW THIS IS GOING TO VIOLATE OUR RULE
THAT WE HAVE TO HAVE WHOLE NUMBERS.
SO WE'LL TREAT THIS [INAUDIBLE] ALGEBRAIC EXPRESSION
AND WE'LL MULTIPLE THE ENTIRE EQUATION BY THE FACTOR OF TWO.
SO THAT WILL GIVE US OUR BALANCED REACTION
FOR ALUMINUM ATOMS OR FOUR MOLES WILL REACT WITH THREE MOLES
OF OXYGEN TO PRODUCE TWO MOLES OF ALUMINUM OXIDE.
ALL RIGHT.
NOW LET'S TRY ANOTHER ONE. ALL RIGHTY.
SO HERE, EVERYTHING ALL OF--
THEY'RE ALL COMPOUNDS WITH MIXED ELEMENTS
SO WE JUST PICK A PLACE TO START,
MY EXPERIENCE I PICK AN ATOM AND I DO BEST,
IF I FIND MYSELF GETTING STUCK, I START OVER FRESH
AND BEGIN WITH A DIFFERENT ELEMENT THAT SEEMS TO BE
ALL IT TAKES TO GET UNSTUCK IN THIS TYPE OF QUESTION.
ALL RIGHTY.
WELL, THE PHOSPHORUSES ARE BOTH ONES, SO WE'LL WAIT THERE.
WE SEE FIVE CHLORINES AND ONE CHLORINE,
SO THAT WOULD LET US KNOW WE SHOULD PUT A FIVE HERE
AND THEN THE OXYGENS, WE HAVE ONE OXYGEN
AND WE HAVE FOUR OXYGENS SO WE'LL PUT A FOUR HERE.
NOW THE MOMENT OF TRUTH,
LET'S LOOK AT THOSE HYDROGENS AND SEE WHAT'S GOING ON.
ON THE LEFT 4X2 WE HAVE EIGHT HYDROGENS
AND THEN WE HAVE THREE HYDROGENS FROM THE PHOSPHORIC ACID
AND FIVE HYDROGENS FROM THE HYDROCHLORIC AND SO VOILA.
OUR REACTION IS BALANCED. OKAY.
LET'S TRY ANOTHER ONE HERE
THE BARIUMS ONE TO ONE LOOKS GREAT,
WE HAVE TWO OXYGENS BUT WE HAVE THREE OVER HERE,
WELL WOULDN'T IT BE SIMPLER IF THERE WAS JUST ONE,
SO I'D SAY WE CAN MULTIPLY THAT BY ONE HALF,
RIGHT 1/2 X 2 WOULD BE ONE BUT OF COURSE,
NOW WE'VE INTRODUCED A FRACTIONAL COEFFICIENT,
SO WE'LL USE THAT SAME STRATEGY
AND MULTIPLY THE ENTIRE EXPRESSION BY TWO
TO GET OUR BALANCED EQUATION SO TWO MOLES
OF BARIUM PEROXIDE PRODUCED TWO MOLES OF BARIUM OXIDE
AND A MOLE OF OXYGEN, OKAY.
AND THEN THERE IS A SEPARATE VIDEO TUTORIAL ON THIS REACTION,
THIS IS A COMBUSTION REACTION
BUT I THINK IT'S WORTH TAKING THE TIME TO DO ONE
HERE AS WELL.
SO REMEMBERING WITH COMBUSTION REACTIONS
WE WANT TO START WITH THE CARBONS
SO WE HAVE FIVE ON THE LEFT SO WE WOULD NEED FIVE
CO2S AND THEN WE HAVE TEN HYDROGENS,
SO WE WOULD NEED FIVE WATERS SO
WE LOOK AT ALL OF THE OXYGENS ON THE RIGHT,
WE HAVE A FIVE AND THEN THERE'S TWO IN EACH CO2
AND THEN WE HAVE A FIVE COEFFICIENT IN FRONT OF WATER,
EACH WATER HAS ONE,
SO THAT LEAVES US WITH FIFTEEN OXYGENS,
OXYGEN IS HIGH MAINTENANCE, RIGHT SO WE'RE BACK TO 15/2
AND ONCE AGAIN WE WILL MULTIPLY THE ENTIRE EXPRESSION
BY TWO TO CLEAR THOSE FRACTIONAL COEFFICIENTS
AND I'LL SIDE THIS UP JUST A WEE BIT MORE
AND WE'LL PUT OUR FINAL ANSWER TO--
OF THE HYDRO CARBONS REACTS WITH 15 OXYGEN TO PRODUCE 10 CO2
AND 10-- OOPS 10 WATERS.
OKAY. SO THAT CONCLUDES OUR VIDEO TUTORIAL
ON BALANCING CHEMICAL REACTIONS,
THIS WOULD BE A GREAT TIME TO WORK A FEW MORE PROBLEMS
TO REINFORCE YOUR UNDERSTANDING.