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fredd [130]
4 years ago
15

Why do we need to balance chemical equations?

Chemistry
2 answers:
Mrrafil [7]4 years ago
8 0
So we know the number of moles of each compound. If we need to know the concentration we must know the number of moles that the compounds react with...

GrogVix [38]4 years ago
6 0
The law of conservation of mass states that matter can neither be created nor destroyed, and it just transforms from one medium to another. Therefore, when a chemical reaction occurs, we need to balance the equation to ensure that the number of particles of the reactants are equal to the number of particles of the products.

Consider the following (unbalanced) equation:
H2 + O2 -> H2O
If this is what actually happens, then where is the second atom of Oxygen?

To ensure that stuff like this doesn't happen, we have to balance equations:
2H2 +O2 -> 2H2O

Now if you count the atoms, you have 4 Hydrogen atoms and 2 Oxygen atoms on both sides, and conservation of mass takes place.
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Because of their different longitude.

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3 years ago
) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react
Tom [10]

<u>Answer:</u> The amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of B_5H_9 with oxygen gas follows:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ

To calculate the amount of heat released for the given amount of B_5H_9(l), we use unitary method, we get:

When 2 moles of B_5H_9(l) reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of B_5H_9(l) will react, the amount of heat released will be = \frac{5259}{2}\times 0.211=554.8kJ

Hence, the amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

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selective breeding

Explanation:

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