Balance each of the following chemical equations: a. K(s)+H2 O(l) ⟶H2(g)+KOH(a q) b. Cr(s)+S8(s)

Balance each of the following chemical equations: a. K(s)+H2...

Balance each of the following chemical equations: a. $\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{KOH}(a q)$ b. $\mathrm{Cr}(s)+\mathrm{S}_{8}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)$ c. $\mathrm{BCl}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{BO}_{3}(a q)+\mathrm{HCl}(a q)$ d. $\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)$ e. $\mathrm{BaCl}_{2}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{NaCl}(a q)$

Balance each of the following chemical equations:
a. $\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{KOH}(a q)$
b. $\mathrm{Cr}(s)+\mathrm{S}_{8}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)$
c. $\mathrm{BCl}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{BO}_{3}(a q)+\mathrm{HCl}(a q)$
d. $\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)$
e. $\mathrm{BaCl}_{2}(a q)+\mathrm{Na}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{NaCl}(a q)$

Key Concepts

Law of Conservation of Mass

This principle states that in a chemical reaction, matter is neither created nor destroyed. Every atom present in the reactants must appear among the products. The law forms the basis for balancing chemical equations, ensuring that the same number of each type of atom is found on both sides of the equation.

Stoichiometry

Stoichiometry involves using the quantitative relationships between reactants and products in a chemical reaction. When balancing chemical equations, stoichiometric coefficients are assigned to compounds to reflect the mole ratios that conserve the number of atoms in the reaction.

Balancing Chemical Equations

Balancing chemical equations is the process of determining the correct coefficients for all reactants and products so that the number of atoms of each element is the same on both sides of the reaction. This practice is essential for accurately representing chemical reactions and performing subsequent calculations involving reactants and products.

Coefficients

Coefficients are the numbers placed in front of chemical formulas in an equation. They indicate the proportion of molecules or moles involved in the reaction, and their proper determination ensures compliance with the law of conservation of mass.

Transcript

00:01 So for this question, we're given an equation and we have to balance it.

00:04 So that means we need the same number of atoms of each element on each side.

00:09 So for this one, we've got one potassium here, one potassium here, so that's good.

00:14 Then we have two hydrogens here, two hydrogens here, and one here.

00:18 So we've got two, and then we have three.

00:21 So since this is h2 only on this side, we know we're going to have to have an even number on our right side to make this doable to get an even number on the left side as well, because we can only have an even number because it's h2.

00:36 So we're going to multiply this by 2.

00:39 So then we have h2 here and 2h here, so that's a total of 4.

00:47 Then we end up with 2 potassium, and multiply that by 2, so we have 2 potassium here.

00:53 Then we also have 2 oxygen, and we're going to need 2 oxygen.

01:01 As well on this side, and that balances it out, so we end up with 2k plus 2h2o, h2 plus 2kh, and that's our balanced equation.

01:19 For part b, we've got cr, cr2, so i'm going to multiply this by 2.

01:25 Then we have s8 and s3.

01:27 So when you have two that are this different, it's really easy to find a common denominator.

01:36 By multiplying by one or the other, right? so we can either say, okay, well, there's eight here, so i'm going to multiply by three here.

01:48 And then we've got 24, and then multiply by 8 here, and we've got 24, right? and then since cr is by itself, we can do 8 times 2.

01:58 That's 16, so this becomes 16 plus 3, s8, and then 8 cr2 is 3 and that's our balanced equation then for part c we have one boron here one bore on here we have 3 chlorine and only 1 here so multiply this by 3 then for our hydrogen we've got 2 here we've got 3 here and 3 here so that's total of 6 so to make this 6 we'll multiply by 3 and then then for our oxygen, we've got three now, multiply by three, and then this is also three.

02:45 So that is a balanced equation right there...

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