SAT Chem balancing

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1

<p>I'm in need of dire help with balancing equations. I know this is easy for a lot of you, but I'm not very good at chemistry, so please help me out. And please... don't give an answer without explaining.</p>

<p>Balance: Cu + HNO3 ---> Cu(NO3)2 + H2O + NO</p>

<p>I know that if I place a 2 infront of HNO3, the hydrogens will be balanced, but that would still leave me with 6 oxygens on one side - 8 oxygens on the other, and 2 nitrogens on one side and 3 nitrogens on the other.</p>

2

<p>Well first off the equation is wrong; NO should be NO2.</p>

<p>Cu + 4HNO3 -> Cu(NO3)2 + 2H2O + 2NO2</p>

<p>As a general rule you do hydrogens and oxygens last when you’re balancing…in this case you’d do nitrogen first. There’s one on the reactants side and three on the products, but if you look at the products, there needs to be an even number of Hs. </p>

<p>So in this case, you’d try 4 as the coefficent for nitric acid. Then balance for your hydrogens to get 2 for H2O, then do oxygens and you’ll get 2 for NO2.</p>

3

<p>You essentially need to play with it. Here’s the solution that a Google search yielded:</p>

<h1>Balance: Cu + HNO3 —> Cu(NO3)2 + H2O + NO</h1>

<p>Cu + 2HNO3—> Cu(NO3)2 + H2
2HNO3—> H2O+N2O5
N2O5 + 3H2-----> 3H2O+2NO</p>

<p>3Cu+ 8HNO3----> 3Cu(NO3)2 + 2NO +4H2O</p>

<p>For such complex problems you may need the Oxidation State Method. <a href=“http://www.mpcfaculty.net/mark_bishop/redox_balance_oxidation.htm[/url]”>http://www.mpcfaculty.net/mark_bishop/redox_balance_oxidation.htm&lt;/a&gt;&lt;/p&gt;