Under the right conditions aluminum will react with chlorine to produce aluminum chloride. 2 Al + 3 Cl2 – 2 AlCl3 How many grams

Question

Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
2 Al + 3 Cl2 – 2 AlCl3
How many grams of aluminum are needed to react completely with 11.727 liters of chlorine?

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Neala 5 years 2021-08-02T20:49:32+00:00 1 Answers 6 views 0

Answers ( )

  1. Eirian
    0
    2021-08-02T20:51:10+00:00
    Reply

    Answer:

    m_{Al}=9.42gAl

    Explanation:

    Hello there!

    In this case, according to the given chemical reaction:

    2 Al + 3 Cl2 –> 2 AlCl3

    Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

    m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2} *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

    Regards!

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