Balance the following equations: (a) The explosion of...

Balance the following equations: (a) The explosion of ammonium nitrate: $\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2}+\mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}$ (b) The spoilage of wine into vinegar: $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+\mathrm{O}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}$ (c) The burning of rocket fuel: $\mathrm{C}_{2} \mathrm{H}_{8} \mathrm{N}_{2}+\mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{N}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

Balance the following equations:
(a) The explosion of ammonium nitrate:
$\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{N}_{2}+\mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}$
(b) The spoilage of wine into vinegar:
$\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}+\mathrm{O}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}$
(c) The burning of rocket fuel:
$\mathrm{C}_{2} \mathrm{H}_{8} \mathrm{N}_{2}+\mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{N}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is the process of adjusting the coefficients of reactants and products to satisfy the conservation of mass. This systematic approach ensures that the number and type of atoms are equal on both sides of the equation, providing a complete and correct representation of the chemical reaction.

Conservation of Mass

This principle dictates that matter is neither created nor destroyed during a chemical reaction. When balancing chemical equations, the number of atoms of each element must be the same on both the reactant and product sides, reflecting the conservation of mass.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It uses the coefficients from the balanced equation to calculate the amounts of substances consumed and produced, ensuring accurate and predictable outcomes in reactions.

Transcript

00:01 I like to follow these guidelines for balancing reactions.

00:05 There's no polyethomic ions really to deal with on both sides.

00:09 So i'm going to start out by balancing, i think, the hydrogens, because they're the only one, it's the only element that appears once on each side of the equation.

00:20 So looking at hydrogen, i've got four on the left, two on the right.

00:25 So i'm going to say that for every one of these ammonium nitrates, i'm going to need two waters.

00:32 Now i can't change either of these coefficients.

00:37 So i'll look at the next easiest substance.

00:42 It seems that i have two nitrogens on the left.

00:46 So i need a coefficient of one right here.

00:49 And that means that all i have left to deal with is this oxygen, which is a little tricky because there's an oxygen right here too.

00:55 So there's a total of three oxygen on the left side and right now there are two and two.

01:05 So this is one of those weird scenarios where one solution is to say well if i had a coefficient of one half in front of the o2 that would bring me to one oxygen right here and now my oxygens would be balanced three on each side.

01:22 But obviously i can't have coefficients of one half.

01:25 So what i'm going to do is multiply through by 2 so that all my coefficients are whole numbers.

01:34 And what that does, if it changes this into a 2, this into a 2, that's a 1, this is a 4, and now my reaction is balanced.

01:58 Next, i'll start with a more complicated compound.

02:04 I'm going to balance these 2 using carbon, because that appears once on each side.

02:11 I can see that one of each molecules will balance the carbons, and then i can start trying to figure out what to do with the rest...

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