Balance each of the following chemical equations. a. K2 SO4(a q)+BaCl2(a q) →BaSO4(s)+KCl(a q) b

step 1 Problem 39 asks us to balance these chemical equations. For the first one, we have two potassium

Balance each of the following chemical equations. a. K2...

Balance each of the following chemical equations. a. $\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{KCl}(a q)$ b. $\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{FeO}(s)+\mathrm{H}_{2}(g)$ c. $\mathrm{NaOH}(a q)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{NaClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ d. $\operatorname{Mg}(s)+\operatorname{Mn}_{2} \mathrm{O}_{3}(s) \rightarrow \mathrm{MgO}(s)+\operatorname{Mn}(s)$ e. $\mathrm{KOH}(s)+\mathrm{KH}_{2} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ f. $\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HNO}_{3}(a q)$ g. $\mathrm{BaO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{O}_{2}(g)$ h. $\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l)$

Balance each of the following chemical equations.
a. $\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{KCl}(a q)$
b. $\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{FeO}(s)+\mathrm{H}_{2}(g)$
c. $\mathrm{NaOH}(a q)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{NaClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
d. $\operatorname{Mg}(s)+\operatorname{Mn}_{2} \mathrm{O}_{3}(s) \rightarrow \mathrm{MgO}(s)+\operatorname{Mn}(s)$
e. $\mathrm{KOH}(s)+\mathrm{KH}_{2} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
f. $\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{HNO}_{3}(a q)$
g. $\mathrm{BaO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{O}_{2}(g)$
h. $\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l)$

Key Concepts

Law of Conservation of Mass

This principle states that mass is neither created nor destroyed in a chemical reaction, meaning that the total number of atoms for each element must remain constant on both sides of the equation. It is the fundamental reason why chemical equations need to be balanced.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions using the mole ratios derived from the balanced equation. It ensures the correct proportional relationships are maintained to satisfy the conservation of mass.

Balancing Chemical Equations

This concept focuses on adjusting the coefficients of reactants and products to achieve an equal number of atoms for each element on both sides of the equation. It is a step-by-step process that reflects the conservation properties inherent in chemical reactions.

Chemical Reaction Notation and Phases

Chemical equations use specific notations to indicate the states of matter of reactants and products, such as solids, liquids, gases, or aqueous solutions. These designations, while not affecting the balancing process directly, provide important context about the physical conditions of the reaction.

Oxidation-Reduction (Redox) Reactions

In some chemical reactions, such as those involving metals and oxygen, there is a transfer of electrons that changes the oxidation states of the elements involved. Recognizing redox processes is important as they often require a careful balancing of both mass and charge, which can add complexity to the balancing process.

Transcript

00:01 Problem 39 asks us to balance these chemical equations.

00:05 For the first one, we have two potassium on the left, so we need to make two potassium on the right.

00:12 So that balances our equation.

00:15 In the second equation, one iron in the left, one iron in the right, two hydrogens on the left, two hydrogen on the right, one oxygen on the left, one oxygen on the right.

00:26 This one is already balanced.

00:30 And then for this next one, we have one sodium on the left, one sodium on the right, one chlorine, on the left, one chlorine on the right, five oxygens on the left, five of oxygens on the right, two hydrogens on the left, two hydrogens on the right.

00:44 This one's already balanced as well.

00:48 This next one we have two manganese on the left, so balance, we need two over here.

00:54 We have three oxygens on the left, so we add three here.

00:59 And now we need to also add three on the left in front of magnesium to balance.

01:06 So for this one, i looked at the number of hydrogen, first and i could see that there are three on the left, two on the right.

01:15 So to make four on both sides, i added a two here and a two here.

01:24 And this also balances our potassiums with three on each side.

01:31 And our oxygens are also balanced.

01:34 So this equation is balanced.

01:39 So for this next one, the number of nitrogens are equal on both sides.

01:48 The number of hydrogens is two over here.

01:51 So put a two over here.