Lesson 12 Balancing Chemical Reactions
Balancing Chemical Reactions
Balancing a chemical reaction equation ensures that the law of conservation of mass is followed.
B12.10 Writing a chemical reaction
A balanced chemical reaction equation shows that the atoms are conserved during a chemical reaction. Coefficients are added to skeleton equations to ensure that the number of atoms of each element on the reactant side equals the number of
atoms of each element on the product side.
Remember, you cannot change the subscripts in a formula to balance an equation. That would alter the identity of the compound. Instead, use coefficients.
Remember, you cannot change the subscripts in a formula to balance an equation. That would alter the identity of the compound. Instead, use coefficients.
For example, given the skeleton equation for the production of water from hydrogen and oxygen,
H2(g) + O2(g) → H2O(l)
you can immediately notice that the oxygen atoms are not balanced. By placing a coefficient of 2 in front of the water, the oxygen atoms will be balanced:
H2(g) + O2(g) → 2H2O(l)
This, however, causes the H atoms to not be balanced:
H2(g) + O2(g) → 2H2O(l)
But by placing a 2 in front of the hydrogen gas, they will be balanced:
2H2(g) + O2(g) → 2H2O(l)
By convention, no 1 is placed in front of the oxygen gas; it is assumed to be there.
You can check your balancing by making a chart that compares the number of atoms on the reactant side to the number of atoms on the product side.
| Entity | Reactants | Products |
|---|---|---|
| H | 4 | 4 |
| O | 2 | 2 |
Watch this video to see a teacher work through this example. https://adlc.wistia.com/medias/5q00dqu3j7
Examples
Work though the following examples to ensure you have strong comprehension of balancing chemical reaction equations. You will be balancing by inspection; so the best place to start is with the element on the far left of the equation and then working your way through all elements in order. Each example has a video to go with it. To play the video, click on the play icon next to the example.
Balance the following skeleton reaction equation:
Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)
Balance by inspection:
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Mg atoms are balanced, as there is 1 Mg on both sides of the equation.
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Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)
Balance by inspection:
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Mg atoms are balanced, as there is 1 Mg on both sides of the equation.
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Cl atoms are balanced, as there are 2 Cl atoms on both sides of the arrow. The balanced chemical reaction equation is
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
| Entity | Reactants | Products |
|---|---|---|
| Mg | 1 | 1 |
| H | 2 | 2 |
| Cl | 2 | 2 |
Tip: If a polyatomic ion stays together as a unit, it can be balanced as one entity instead of looking at each individual atom within the polyatomic.
Balance the following skeleton reaction equation:
AlI3(aq) + AgNO3(aq) → AgI(s) + Al(NO3)3(aq)
Balance by inspection:
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Balance the following skeleton reaction equation:
AlI3(aq) + AgNO3(aq) → AgI(s) + Al(NO3)3(aq)
Balance by inspection:
«math»«mtable»«mtr»«mtd»«msub»«mi»AlI«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»AgNO«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»AgI«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»Al«/mi»«mo»(«/mo»«msub»«mi»NO«/mi»«mn»3«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable»«mtr»«mtd»«msub»«mi»AlI«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»AgNO«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»AgI«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»Al«/mi»«mo»(«/mo»«msub»«mi»NO«/mi»«mn»3«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
I atoms are not balanced, as there are 3 I atoms on the reactant side and only 1 I atom on the product side. Add a coefficient of 3 in front of AgI.
«math»«mtable»«mtr»«mtd»«msub»«mi»AlI«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»AgNO«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«mi»AgI«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»Al«/mi»«mo»(«/mo»«msub»«mi»NO«/mi»«mn»3«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
Ag atoms are not balanced, as there is 1 Ag atom on the reactant side and 3 Ag atoms on the product side. Add a coefficient of 3 in front of the AgNO3.
«math»«mtable»«mtr»«mtd»«msub»«mi»AlI«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«msub»«mi»AgNO«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«mi»AgI«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»Al«/mi»«mo»(«/mo»«msub»«mi»NO«/mi»«mn»3«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
NO3– ion is balanced, as there are 3 NO3– ions on either side of the equation.
AlI3(aq) + 3AgNO3(aq) → 3AgI(s) + Al(NO3)3(aq)
| Entity | Reactants | Products |
|---|---|---|
| Al | 1 | 1 |
| I |
3 | 3 |
| Ag | 3 | 3 |
| NO3 | 3 | 3 |
Balance the following skeleton reaction equation:
Fe(s) + Br2(l) → FeBr3(s)
Balance by inspection:
«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
Fe atoms are balanced, as there is 1 Fe atom on both sides of the reaction.
«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
Fe(s) + Br2(l) → FeBr3(s)
Balance by inspection:
«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
Fe atoms are balanced, as there is 1 Fe atom on both sides of the reaction.
«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»2«/mn»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
Fe atoms are no longer balanced, as there is 1 atom on the reactant side and 2 atoms on the product side. Add a coefficient of 2 in front of Fe.
2Fe(s) + 3Br2(l) → 2FeBr3(s)
| Entity | Reactants | Products |
|---|---|---|
| Fe |
2 | 2 |
| Br |
6 | 6 |
Balance the following skeleton reaction equation:
C4H10(g) + O2(g) → CO2(g) + H2O(g)
Watch this video to learn how to balance hydrocarbon combustions by inspection.
C4H10(g) + O2(g) → CO2(g) + H2O(g)
Watch this video to learn how to balance hydrocarbon combustions by inspection.
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Read This
Please read pages 87 to 88 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the technique for balancing chemical
reaction equations. Remember, if you have any questions or you do not understand something, ask your teacher! Practice Questions
Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice
question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.Balance the following skeleton reaction equations.
- __Fe(s) + __S8(s) → __FeS(s)
8Fe(s) + S8(s) → 8FeS(s) - __KClO3(s) → __KCl(s) + __O2(g)
2KClO3(s) → 2KCl(s) + 3O2(g) - __H3PO4(aq) + __KOH(aq) → __K3PO4(aq) + __HOH(l)
H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3HOH(l) - __Fe2O3(s) + __C(s) → __Fe(s) + __CO(g)
Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g) - __K2O(s) + __H2O(l) → __KOH(aq)
K2O(s) + H2O(l) → 2KOH(aq) - __C6H6(l) + __O2(g) → __CO2(g) + __H2O(g)
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(g) - __K3PO4(aq) + __MgCl2(aq) → __Mg3(PO4)2(s) + __KCl(aq)
2K3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s) + 6KCl(aq) - __SO2(g) + __H2O(l) → __H2SO3(aq)
SO2(g) + H2O(l) → H2SO3(aq) - __LiCl(aq) + __F2(g) → __LiF(aq) + __Cl2(g)
2LiCl(aq) + F2(g) → 2LiF(aq) + Cl2(g) - __NiBr2(aq) + __Al(s) → __AlBr3(aq) + __Ni(s)
3NiBr2(aq) + 2Al(s) → 2AlBr3(aq) + 3Ni(s)