Balance the following chemical equations. MnO2(s)+CO(g) →Mn2 O3(a q)+CO2(g) Al(s)+H2 SO4(a q) →A

step 1 Hi everyone so in this question you have to balance the equation. So first one is MNO2 solid plu

Balance the following chemical equations. MnO2(s)+CO(g) →Mn2...

Balance the following chemical equations. $\mathrm{MnO}_{2}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2} \mathrm{O}_{3}(a q)+\mathrm{CO}_{2}(g)$ $\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)$ $\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ $\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{I}_{2}(g)$ $\mathrm{KOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

Balance the following chemical equations.
$\mathrm{MnO}_{2}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2} \mathrm{O}_{3}(a q)+\mathrm{CO}_{2}(g)$
$\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)$
$\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
$\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{I}_{2}(g)$
$\mathrm{KOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

Key Concepts

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It helps ensure that the number of atoms for each element is conserved by determining the appropriate coefficients when balancing chemical equations.

Law of Conservation of Mass

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction, meaning that the total number of atoms of each element must remain the same on both sides of the equation. This principle is fundamental when balancing chemical equations.

Coefficient Adjustment

Coefficient adjustment is the process of placing the correct numerical multipliers in front of compounds or elements in a chemical equation. This method is used to ensure that the number of atoms for each element is equal on both the reactant and product sides of the equation.

Systematic Balancing Strategy

A systematic balancing strategy involves selecting an order in which to balance the elements, typically starting with the most complex molecule or the element that appears in a single compound on each side. This strategy helps organize the balancing process and reduces errors during the adjustment of coefficients.

Transcript

00:01 Hi everyone so in this question you have to balance the equation.

00:03 So first one is mno2 solid plus carbon monoxide gas gas mn2 amn2 o3 acquies plus co2 gas now in this question if you see properly carbon item they are balanced, okay? then auction atoms, they are not balanced.

00:46 But if you see properly, at right hand side there are 2 mm.

00:50 So, of course you are right, a 2 at left hand side.

00:57 So what will get is toys, then mno2 plus co gas gives mn2o3 aquas, plus co2 gas.

01:21 Now, this is a balanced equation.

01:25 At, you know, reactants, i'll get different items like manganese, oxygen and carbon.

01:31 So, manganese are two.

01:33 Then oxygen, so two -twos are four plus one for you.

01:37 Okay, then carbon two.

01:39 At product side, manganese are two.

01:43 Then carbon, one and auction, three plus.

01:47 H2 5.

01:48 So the equation is balanced.

01:51 Then second equation, aluminium solid plus h2so4 acquies gives altaso4 thrice plus acquies plus h2 gas.

02:23 So the balance equation is the balance equation is the balance equation is as follows.

02:43 Now, in this case also, you can find that is, there are two aluminium at right -hand side.

02:54 So you have to make two aluminium at left -hand side, two aluminium at left -hand side plus.

03:09 At right -hand side, there are three so -4 units, and that's why you have to multiply this h -2 -sopor -a -lept -a -side by 3.

03:18 So 3 into h2 s.

03:26 S.

03:26 4.

03:26 Equius gives al to s.

03:33 4 thrice equius plus...

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