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Postsecondaire • 1a

Why is option 4 not right

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Chemistry
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  • Explication d'Alloprof

    Explication d'Alloprof

    Cette explication a été donnée par un membre de l'équipe d'Alloprof.

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    Équipe Alloprof • 1a December 2024 modifié

    Hi EfficientEagle2708, thanks for your question !

    English is not my primary language but i'll do my best to assist you :)

    So the fourth statement is indeed incorrect, because the coefficient always applies for molar ratios.

    It is because the mass isn't proportional in a reaction. For example, 100g of dihydrogen \(H_2\) has way more particles than 100g of difluor \(F_2\). You can determine that by using the formula :

    $$ n = \frac{m}{M}$$

    For the dihydrogen :

    $$ n = \frac{100 g}{2,02 g/mol} = 49,5 mol$$


    For the difluor :

    $$ n = \frac{100 g}{38,00 g/mol} = 2.63 mol$$


    But, when you say that you have 1mol of dihydrogen on one side and 1mol of difluor on another side, you have exactly \(6.022.10^{23}\) molecules for each side. And the coefficients will multiply this ratio and guarantee that all of our calculation stays proportionals.


    I hope my explanation is helping you, feel free to write again if you have any other questions !

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