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Example 1: Solid potassium nitrate decomposes when heated to produce solid potassiumm nitrite and oxygen gas.

KNO

_{3}(s) ---> KNO_{2}(s) + O_{2}(g)Step 1: balance K first

It already is. So far not to bad.

Step 2: balance N next

It already is. Boy this is easy.

Step 3: finally, balance O - not so easy, but not too hard.

You have 3 O atoms on the left and 4 O's on the right. How can you balance the O's w/o changing the coefficients of 1 in front of the KNO

_{3}and KNO_{2}? (NOTE: you don't write a 1 when it is the subscript in a formula or the coefficient in front of a formula in an equation.) What you need to realize is that the 3 O's in the KNO_{3}and 2 O's in the KNO_{2}shouldn't be changed directly since that would throw K and N out of balance. So what you actually have to ask is "How can I balance the 3 O's on the left starting with 2 O's on the right (in the KNO_{2})?". Write it like this: KNO_{3}(s) ---> KNO_{2}(s) + O_{2}(g) 3 O's 2 O's + "How many O's do I need from O_{2}to give a total of 3?" The answer of course is you need only 1 O atom from O_{2}to add to the 2 O's you already have to give a total of 3 O's on the right. How can you get 1 O atom from an O_{2}? If you take 1/2 of an O_{2}molecule that will give you 1 O atom. You use a coefficient of 1/2 in front of the O_{2}. This will give you the 1 O that you need to balance the O atoms. So at this point you would have the following: KNO_{3}(s) ---> KNO_{2}(s) + 1/2 O_{2}(g) NOTE: Using a fraction isn't that difficult. You needed ONE O atom. That's what goes in the numerator. The subscript for the O_{2}is a TWO. That's what's in the denominator. If it had been an O_{3}you would have used a fraction with a 3 in the denominator. However, remember one of the requirements is to have whole number coefficients (Although there are times we do see fractions. It depends on what you are trying to use the equation for, but let's not get into that right now.) To get whole numbers you simply multiply the whole equation by 2. This will give the following: 2 KNO_{3}(s) ---> 2 KNO_{2}(s) + O_{2}(g) Don't forget there is a "1" in front of the O_{2}.Step 4: Check to make sure everything is balanced.

__Left____Right__2 K 2 K 2 N 2 N 6 O 6 O So everything is balanced. This wasn't too tough, was it?

Example 2: Balance the following eqn.

PCl

_{3}+ H_{2}O ---> H_{3}PO_{3}+ HCl Hint: Start with the Cl. Answer: PCl_{3}+ 3 H_{2}O ---> H_{3}PO_{3}+ 3 HClExample 3: Balance the following eqn.

CaC

_{2}+ H_{2}O ---> Ca(OH)_{2}+ C_{2}H_{2}Answer: CaC_{2}+ 2 H_{2}O ---> Ca(OH)_{2}+ C_{2}H_{2}Example 4: Balance the following eqn.

C

_{2}H_{2}+ O_{2}---> CO_{2}+ H_{2}O Hint: Start with the C. You may want to use a fraction to help you balance the oxygens and save the O atoms until last. What fraction do you need in front of the O_{2}? Where do you think the 5 goes in the fraction (you need 5 O atoms)? What goes in the denominator (see the subscript of 2)? Answer: 2 C_{2}H_{2}+ 5 O_{2}---> 4 CO_{2}+ 2 H_{2}OExample 5: Balance the following eqn.

CaO + H

_{3}PO_{4}---> Ca_{3}(PO_{4})_{2}+ H_{2}O Hint: Balance the PO_{4}as a unit since it appears in that form on both sides of the arrow. This is the PO_{4}^{3-}polyatomic anion. Answer: 3 CaO + 2 H_{3}PO_{4}---> Ca_{3}(PO_{4})_{2}+ 3 H_{2}O

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