Balancing Chemical Equations Class10:How to balance chemical reactons
Balancing chemical equations class 10 involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical equation must obey the law of conservation of mass and the law of constant proportions, i.e. the same number of atoms of each element must exist on the reactant side and the product side of the equation.
Balancing Chemical Equations Class10
Two quick and easy methods of balancing a chemical equation are discussed in this article. The first method is the traditional balancing method and the second one is the algebraic balancing method.
The Key to Balancing Chemical Equations Class10:
Related Terminology Balancing chemical equations class 10 involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical
Chemical Equation(Balancing Chemical Equations Class10):-A chemical equation is a symbolic representation of a chemical reaction in which the reactants and products are denoted by their respective chemical formulae.
An example of a chemical equation is 2H2 + O2 → 2H2O which describes the reaction between hydrogen and oxygen to form water
The reactant side is the part of the chemical equation to the left of the ‘→’ symbol whereas the product side is the part to the right of the arrow symbol.
Example of Balancing a Chemical Equation
P4O10 + H2O → H3PO4
Balancing Chemical Equations Class10
First, let’s look at the element that appears least often. Notice that oxygen occurs twice on the left-hand side, so that is not a good element to start out with. We could either start with phosphorus or hydrogen, so let’s start with phosphorus. There are four atoms of phosphorus on the left-hand side, but only one on the right-hand side. So, we can put the coefficient of 4 on the molecule that has phosphorous on the right-hand side to balance them out.
P4O10 + H2O → 4 H3PO4
Balancing Chemical Equations Class10
Now we can check hydrogen. We still want to avoid balancing oxygen,(Balancing Chemical Equations Class10) because it occurs in more than one molecule on the left-hand side. It is easiest to start with molecules that only appear once on each side. So, there are two molecules of hydrogen on the left-hand side and twelve on the right-hand side (notice that there are three per molecule of H3PO4, and we have four molecules). So, to balance those out, we have to put a six in front of H2O on the left.
P4O10 + 6 H2O → 4 H3PO4
At this point, we can check the oxygens to see if they balance. On the left, we have ten atoms of oxygen from P4O10 and six from H2O for a total of 16. On the right, we have 16 as well (four per molecule, with four molecules). So, (Balancing Chemical Equations Class10)oxygen is already balanced. This gives us the final balanced equation of
P4O10 + 6 H2O → 4 H3PO4
Balancing Chemical Equations Class 10: Practice Problems
Try to balance these ten equations on your own, then check the answers below. They range in difficulty level, so don’t get discouraged if some of them seem too hard. Just remember to start with the element that shows up the least, and proceed from there. The best way to approach these problems is slowly and systematically. Looking at everything at once can easily get overwhelming. Good luck!
- CO2 + H2O → C6H12O6 + O2
- SiCl4 + H2O → H4SiO4 + HCl
- Al + HCl → AlCl3 + H2
- Na2CO3 + HCl → NaCl + H2O + CO2
- C7H6O2 + O2 → CO2 + H2O
- Fe2(SO4)3 + KOH → K2SO4 + Fe(OH)3
- Ca3(PO4)2 + SiO2 → P4O10 + CaSiO3
- KClO3 → KClO4 + KCl
- Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4
- H2SO4 + HI → H2S + I2 + H2O
Complete Solutions:Balancing Chemical Equations Class 10
1)6CO2 + 6H2O → C6H12O6 + 6O2
2)SiCl4 + 4H2O → H4SiO4 + 4HCl
3)2Al + 6HCl → 2AlCl3 + 3H2
4)Na2CO3 + 2HCl → 2NaCl + H2O + CO2
5)2C7H6O2 + 15O2 → 14CO2 + 6H2O
6)Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3-
7)2Ca3(PO4)2 + 6SiO2 → P4O10 + 6CaSiO3
8)4KClO3 → 3KClO4 + KCl
9)Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3 + 3CaSO4
10)H2SO4 + 8HI → H2S + 4I2 + 4H2O
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