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Ber [7]
3 years ago
13

How many grams are in 4.65 mol of AI(NO2)3

Chemistry
1 answer:
nirvana33 [79]3 years ago
6 0
To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ = 
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃

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Answer:

The total amount of heat released  is  68.7 kJ

Explanation:

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Total heat released = -68.7 kJ

Note that the "negative sign" simply indicates heat released, therefore no need to put it in the answer.

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