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Balance the following reaction equation. What is the sum of the coefficients of the reactants and products?
__COCl2 + __H2O --> __CO2 + __HCl
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1 COCl2 + __H2O --> 1 CO2 + __HCl
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Balance the carbons first.
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1 COCl2 + __H2O --> 1 CO2 + 2 HCl
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Next, balance the chlorines. (I am leaving the 1 coefficients in for emphasis.)
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1 COCl2 + 1 H2O --> 1 CO2 + 2 HCl
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Now Balance the hydrogens. Notice how the oxygens take care of themselves!
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1 + 1 + 1 + 2 = 5
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Finally, add up the coefficients!
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Balance the following reaction equation. What is the sum of the coefficients of the reactants and products?
__Al4C3 + __HCl --> __AlCl3 + __CH4
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1 Al4C3 + __HCl --> 4 AlCl3 + __CH4
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Balance the aluminum first.
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Al4C3 + 12 HCl --> 4 AlCl3 + __CH4
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Next, balance the chlorine.
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Al4C3 + 12 HCl --> 4 AlCl3 + 3 CH4
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Now balance the carbon. Once again, one element (hyrdogen) took care of itself!
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1 + 12 + 4 + 3 = 20
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Finally, add the coefficients.
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Balance the following reaction equation. What is the sum of the coefficients of the reactants and products?
__Fe2O3 + __CO --> __CO2 + __Fe
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__Fe2O3 + 3 CO --> 3 CO2 + __Fe
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Balance the carbons first such that we get an even number of oxygens. The number of oxygens must be even since the only place oxygen appears as a product is in CO2!
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1 Fe2O3 + 3 CO --> 3 CO2 + __Fe
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Next, let's balance the oxygens. Normally it is nice to save oxygen for the end. But in this reaction, w have a pure element. Always save those for last - even after oxygen and hydrogen!
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Fe2O3 + 3 CO --> 3 CO2 + 2 Fe
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Lastly, balance the iron.
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1 + 3 + 3 + 2 = 9
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Finally, add the coefficients.
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Balance the following reaction equation. What is the sum of the coefficients of the reactants and products?
__C4H10O + __O2 --> __CO2 + __H2O
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1 C4H10O + __O2 --> 4 CO2 + __H2O
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First, balance the carbons using carbon dioxide.
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C4H10O + __O2 --> 4 CO2 + 5 H2O
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Next, balance the hydrogens with water.
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C4H10O + 6 O2 --> 4 CO2 + 5 H2O
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Lastly, balance the oxygens.
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1 + 6 + 4 + 5 = 16
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Now add the coefficients.
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Note: make sure that when you are done that the coefficients are the smallest integers which actually do the job of properly balancing the reaction!
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Balance the following reaction equation. What is the sum of the coefficients of the reactants and products?
__Mg + __FeCl3 --> __MgCl2 + __FeCl2
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__Mg + 2 FeCl3 --> __MgCl2 + 2 FeCl2
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First, balance the iron. But keep in mind that we need an even number of chlorines because of the formulae on the product side!
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__Mg + 2 FeCl3 --> 1 MgCl2 + 2 FeCl2
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Next, finish balancing the chlorines.
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1 Mg + 2 FeCl3 --> MgCl2 + 2 FeCl2
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Now balance the magnesium.
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1 + 2 + 1 + 2 = 6
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And the sum of the coefficients is . . .
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Balance the following reaction equation. What is the sum of the coefficients of the reactants and products?
__CuCl2 + __KI --> __CuI + __KCl + __I2
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2 CuCl2 + __KI --> __CuI + __KCl + __I2
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Iodine showing up in multiple places on teh product side is what makes this problem a challange. Let's try picking coefficents that give us four iodine atoms on the product side. Keeping that in mind, let's try a two in front of the CuCl2.
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2 CuCl2 + __KI --> __CuI + 4 KCl + __I2
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Now let's balance the chlorine . . .
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2 CuCl2 + 4 KI --> __CuI + 4 KCl + __I2
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and then the potasium.
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2 CuCl2 + 4 KI --> 2 CuI + 4 KCl + __I2
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There are the four iodines I wanted. Now let's balance copper.
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2 CuCl2 + 4 KI --> 2 CuI + 4 KCl + 1 I2
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The reaction equation will now be balanced with the proper ceofficient on the I2.
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2 + 4 + 2 + 4 + 1 = 13
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And here is the sum.
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Note: This is an example of aa redox reaction because we see that copper is reducing and iodine is oxidizing. These reactions are easier to balance systematically by breaking up the reaction into half reactions. We will learn more about this process later on.
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