In a reaction, 5.3g of Sodium Carbonate reacted with 6g of ethanoic acid. The products were 2.2g of CO2, 0.9g of water and 8.2g
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Explanation:
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In this case, for this stoichiometry-based problem, it is firstly necessary to realize that the decomposition of ammonium dichromate is given by:
Thus, since the mole ratio between ammonium dichromate and the gaseous nitrogen (molar mass = 28.02 g/mol) is 1:1, we can compute the produced mass of the latter via stoichiometry as shown below:
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<u>Answer:</u>
<u>Balancin</u>g is making the number of atoms of each element same on both the sides (reactant and product side).
To find the number of atoms of each element we multiply coefficient and the subscript
For example contains
5 × 1 = 5,Ca atoms and
5 × 2 = 10, Cl atoms
If there is a bracket in the chemical formula
For example we multiply coefficient
subscript
number outside the bracket.......... to find the number of atoms
(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )
So
3 × 3 = 9, Ca atoms
3 × 1 × 2 = 6, P atoms
3 × 4 × 2 = 24, O atoms are present.
So
Let us balance the equation given
(Unbalanced)
Reactant side - Number of atoms of each element - Product side
1 - Sb - 1
2 - Cl - 3
So we have on the product side odd number of Cl to convert to even number, multiply by 2
The equation changes to
Reactant side - Number of atoms of each element - Product side
1 - Sb - 2
2 - Cl - 6
Multiplying Sb by 2 and
The equation is balanced now!!!!
Reactant side - Number of atoms of each element - Product side
2 - Sb - 2
6 - Cl - 6
Balanced.