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Transcript of Balancing Equations
4 hydrogens and
4 oxygens Here are some new classifications. Aluminum Sodium Aluminum oxide Sodium oxide So... Aluminum Sodium 1 Aluminum reacts with 1 sodium and 2 oxygens to form 2 aluminum, 3 oxygens and 1 sodium This is a single replacement reaction. If you start with oxygen there is 1 oxygen on the reactant side and 3 oxygens on the product side. So you need 3 sodium oxides to equal the 3 oxygens in the product side. Sodium oxide Aluminum oxide With 1 aluminum oxide that makes 2 aluminums so there needs to be 2 aluminums on the reactant side. Aluminum And with 3 sodium oxides there are 6 sodiums so there needs to be 6 sodiums on the product side. Sodium Aluminum Sodium oxide Aluminum oxide Sodium 2 alumiums... 6 sodiums and... 3 oxygens... on the reactant side and ends with... on the product side. on the reactant side... to balance 2 aluminums, 3 oxygens and... 6 sodiums on the product side. 2 aluminums react with 3 sodium oxides to form 1 aluminum oxide and 6 sodium atoms. Lets try another one... Lets try a harder one. sodium carbonate aluminum hydroxide sodium hydroxide aluminum carbonate sodium carbonate aluminum hydroxide sodium hydroxide aluminum carbonate 2 sodiums
3 oxygens 1 aluminum
3 hydrogens 1 sodium
1 hydrogen 2 aluminums
9 oxygens In balancing equations with polyatomic ions it is best to keep them together instead of separating them. This is only possible if the polyatomic ions stay together on both sides of the reaction. So lets recount what we have... 2 sodium
1 carbonate 1 aluminum
3 hydroxides 1 sodium
1 hydroxide 2 aluminums
3 carbonates Our reaction looks like this... This is a double replacement reaction The stradegy to balance double replacement reactions is start with one ion and move on to the next one. aluminum hydroxide sodium carbonate aluminum carbonate sodium hydroxide 1 sodium on the
reactant side 1 sodium on the
product side 1 carbonate on the
reactant side 3 carbonates in the
product side so we need 3 carbonates in the reactant side. by changing sodium carbonate to 3 we need to change sodium hydroxide to 3. checking aluminums there is 1 aluminum on the reactant side... so we need 2 aluminum hydroxides to balance out the aluminums by changing the aluminum hydroxides by 2 it makes 6 hydroxides on the reactant side. So we need to change 3 to a 6. Checking carbonates there is 3 carbonates on the reactant side and there are 3 carbonates on the product side. Everything is balanced. and 2 aluminums on the product side So this is what it looks like all balanced. 6 sodiums
6 hydroxides 2 aluminums
3 carbonates 6 sodiums
Carbon dioxide Water Oxygen gas Lets try one last hard one. This will be a combustion reaction. 4 carbons
10 hydrogens 2 oxygens 1 carbon
2 oxygens 1 oxygen
2 hydrogens Again with combustion balance oxygen last. 4 carbons so... 4 carbon dioxides to create 4 carbons... 10 hydrogens 5 waters to create 10 hydrogens 4 carbon dioxides create 8 oxygens 5 waters creates 5 oxygens 8 + 5 = 13 oxygens on the product side. With oxygen gas, it is not possible to get an odd number of oxygens unless we use a decimal. So we have to double all our numbers because doubleling an odd number creates an even number. Now we have... 2 butanes... 8 carbon dioxides 10 waters 16 oxygens 10 oxygens Now lets recount the oxygens. 16 + 10 = 26 oxygens in the product side. 26/2 = 13 so we need 13 oxygen molecules This is what the balanced equation looks like now. 2 butanes 13 oxygens 8 carbon dioxides 10 water And remember all reactions are present in a larger scale... butane oxygen carbon dioxide water By balancing it is possible to know the ratios between reactants and products. But lets save that for another day... Lets look at what we have. Balancing Chemical Equations by Dalton Ying equations? Lets summarize what
we've learned... In balancing equations the different number of molecules are changed until there are an equal number of atoms on each side of the equation. We change the number of
molecules we have. Without picture the reaction
would be written like this... 1 Methane + 2 oxygens 1 carbon dioxide and 2 water molecules. The number in front of each molecule
is known as coefficient. Without pictures again the reaction would be written like this... 2 aluminums + 3 sodium oxides 1 aluminum oxide + 6 sodiums And without pictures... 3 sodium carbonates +
2 alumium hydroxides 6 sodium hydroxides +
1 aluminum chloride Lastly, without pictures... 2 butanes + 13 oxygens 8 carbon dioxides + 10 waters That number is known as coefficient. There has to be the same number of atoms on each side of the equation. Now lets balance one ion at a time. This makes it possible to calculate how much is used and created.