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anonymous

  • one year ago

How many grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal, if the density of aluminum is 2.70 g/mL? Show all steps of your calculation as well as the final answer. AlCl3 → Al + Cl2 @Photon336 @aaronq @welshfella @dan815

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  1. welshfella
    • one year ago
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    first balance the chemical equation 2 Al Cl3 --> 2 Al + 3 Cl2

  2. welshfella
    • one year ago
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    next work out the mass of the aluminum = 78.3 * 2.7 now from the equation use the relative atomic masses of the elements to find out how much ALCl3 is needed to form 78.3*2.7 grams of Al.

  3. welshfella
    • one year ago
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    Al = 26.982 Cl = 35.45 so from the equation:- 2*26.982 g Al comes from 2*26.982 + 6*35.45 g of AlCl3 next work out how much AlCl3 comes from 1 g of AL then multiply that result by 78.3*2.7

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