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aniked [119]
3 years ago
12

How many moles of aluminum oxide would form if 12.5 moles of aluminum burned​

Chemistry
1 answer:
jarptica [38.1K]3 years ago
7 0
<h3>Al + O2 -> Al2O3</h3>

Balance it:

<h3>2Al + 3O2 -> 2Al2O3</h3><h3 />

So you need 2 Al and 3 O2 to make 2 Al2O3 (aluminum oxide).

I'm going to assume you have all the O2 you need.

Since 2 mols of Al is needed to make 2 mols of the product, it's a 1:1 ratio. You get as much aluminum oxide for as much aluminum you burn.

So 12.5 mols if there is not a lack of the O2.

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How much heat is needed to raise the temperature of 100g of an iron from 25°C to
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382.5J

Explanation:

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What is the ph of a soft drink in which the major buffer ingredients are 6.6 g of nah2po4 and 8.0 g of na2hpo4 per 355 ml of sol
Alex777 [14]
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
                                                   = 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
  = 0.055/ ( 355 ×10^-3)
  = 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)

1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
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 [HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
     = 0.0563/(355×10^-3)
     =  0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation 
Ka of H2PO4- = 6.20 × 10^-8
 [H+] =Ka*([H2PO4-]/[HPO4(2-)]
        = (6.20 ×10^-8)×(0.155/0.1586)
        = 6.059 ×10^-8 M
pH = - log[H+]
     = - log (6.059×10^-8)
     = 7.218

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