![]() Don't be discouraged though, we're actually quite close! To do that, lets add one more molecule of HNO3 to get:Ĭu + HNO 3 = 4 Cu(NO 3) 2 + NO 2 + 2 H 2OĮven though we've now balanced H, we need to balance N again □. As you change coefficients, update the table with the latest atom counts.Īs we predicted, and as can be seen in the table above, we still need to balance H and O. Hopefully you can see that O will become unbalanced as a result, and since like before, 3 is not divisible by 2, multiple steps to balance hydrogen will be needed, so there is more work ahead, but we've got this □:Ĭu + HNO 3 = 3 Cu(NO 3) 2 + NO 2 + 2 H 2Oįor more complex equations, keeping a table like this can make it easier to keep track of your progress. Since H is the only remaining unbalanced element, in the next step, we will put a 2 in front of H2O to try to balance it. To balance the given equation, put a 3 in front of the HNO 3 on the left hand side to get 3 N's on both sides:Īs a side-effect, O was also balanced with 9 on each side. This equation is not balanced because there are an unequal amount of H's, N's and O's on both sides of the equation. ![]() However, the left hand side has 1 H, 1 N, and 3 O's and the right hand side has 2 H's, 3 N's and 9 O's. The left hand side has 1 Cu and the right hand side has 1 Cu, so the Cu atoms are balanced. ![]() Now that we've hopefully gotten the hang of balancing simple chemical equations, lets move onto something a bit more complex, such as the reaction of Copper (Cu) and Nitric Acid (HNO3) into Copper(II) Nitrate (Cu(NO3)2) + Nitrogen Dioxide (NO2) + Water (H2O):
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