Balance each of the following chemical equations. a. Cl2(g)+KBr(a q) →Br2(l)+KCl(a q) b. Cr(s)+O

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

Balance each of the following chemical equations. a. $\mathrm{Cl}_{2}(g)+\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{KCl}(a q)$ b. $\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)$ c. $\mathrm{P}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{PH}_{3}(g)$ d. $\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)$ e. $\mathrm{PCl}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HCl}(a q)$ $\mathrm{f} \cdot \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g)$ g. $\mathrm{C}_{7} \mathrm{H}_{16}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ h. $\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$

Balance each of the following chemical equations.
a. $\mathrm{Cl}_{2}(g)+\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{KCl}(a q)$
b. $\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)$
c. $\mathrm{P}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{PH}_{3}(g)$ d. $\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)$
e. $\mathrm{PCl}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HCl}(a q)$
$\mathrm{f} \cdot \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g)$
g. $\mathrm{C}_{7} \mathrm{H}_{16}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
h. $\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$

Key Concepts

Law_of_Conservation_of_Mass

This principle asserts that matter is neither created nor destroyed during a chemical reaction. In practice, this means that the number and type of atoms must be the same on both sides of a chemical equation, ensuring that the mass remains constant throughout the reaction.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using mole ratios derived from the balanced equation to calculate the amounts of substances consumed and produced, and is fundamental to understanding chemical processes.

Balancing_Chemical_Equations

Balancing chemical equations is the process of adjusting the coefficients of the reactants and products to ensure that the number of atoms for each element is equal on both sides of the equation. This reflects the law of conservation of mass and is a critical skill in chemistry.

Coefficient_Adjustment

Coefficients are the numerical multipliers placed before compounds in a chemical equation. Adjusting these coefficients correctly is key to achieving a balanced equation, as they indicate the mole ratios in which substances react and form, without changing the chemical formulas themselves.

Transcript

0:00 Hi everyone.

00:01 So in this question you have to balance the various equations.

00:05 So first equation is chlorine gas plus kbr acquies gives br2 liquid plus kcl acquies.

00:26 Kcl acquies.

00:28 So the balance equation is balance equation is.

00:41 In order to write the balance equation first write the skeleton equation so cl2 gas plus kbr equius equals br2 liquid plus kcl acquies now there are two chlorines at lhs and rhs only one chlorine so multiply kcl by 2 then there is only one br at lhs and 2 bromins at rhs, so multiply kbr by 2 in order to balance the bromine.

01:27 Now you have to check whether they are balanced or not.

01:30 So there are 3 atoms.

01:32 One is chlorine, potassium and bromine.

01:34 Chlorine 2 on both sides, that is lhs and rhs.

01:39 Podation 2 at lhs as well as rhs, bromine 2 at lhs as well as rhs.

01:45 So the equation is balance.

01:47 Then we will proceed for second equation, that is b equation.

01:51 Crs plus o2 gas gives cr2 o3 gas, o3 gas.

02:06 Now, in order to balance this equation, so what you have to do is, you know, the balance equation is, balance equation is, now, you know, options are 3 over here, and here 2 at lhs.

02:29 So, in order to balance the equation, first write the skeleton equation.

02:35 So plus for two gas, gears, cr2 -o -3 gas.

02:42 Now, at lhs there are two auction atoms, but at rhs there are three auction atoms.

02:47 So multiply this lhs auction atoms by three.

02:50 Then rhs three auction atoms are there, so multiply that by two.

02:55 So we'll get balanced.

02:57 Now chromium atoms they are unbalanced.

03:00 So four chromium atoms are present at rhs and only one chromium present at lhs and therefore multiply the chromium atom at lhs by 4 so you will get the balance equation.

03:12 So just verify you know there are two atoms one is chromium and oxygen.

03:16 So chromium are 4 at lhs as well as rhs then option atoms are 6 at lhs as well as rhs.

03:24 Then we will go for next equation that is c equation.

03:31 So p4 solid plus h2 gas gives phosphine gas.

03:42 Now the balance equation balance equation can be written as follows.

03:55 So in order to write the balance equation first as usual, write the skeleton equation.

04:01 Write the skeleton equation.

04:07 Now, you know, phosphorus, they are 4 at lhs but 1 phosphorus at rhs, so multiply this ph 3 by 4.

04:15 Now 4 3s are 12, 12 hydrogen items at rhs but only 2 hydrogen atom at lhs.

04:22 So multiply this hydrogen molecule by 6.

04:26 Now everything is balanced.

04:29 Now we proceed for d reaction.

04:33 Aluminum solid plus h2 s .o4 equious gives l2 s .o .4 thrice equius plus h2 gas.

04:54 Now the balance equation is balanced equation is, you know, in all to write the balanced equation, first of all right the balanced equation, first of all right, write the skeleton equation.

05:13 S.

05:13 4.

05:14 E .s.

05:16 G.

05:18 L2.

05:29 S .o .4.

05:30 Thrice.

05:32 Equius plus hydrogen gas.

05:37 Now, you know, in order to balance this, what you have to do? you know, so4, there are only one at lhs, but three at rhs.

05:49 So, multiply at lhs by three.

05:53 Now, once you do multiply at 3, 3 -2s are 6 hydrogen atoms at lhs but only 2 hydrogen atom at rhs so multiply the hydrogen atom at rhs by 3...