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Furkat [3]
3 years ago
12

3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent wou

ld remain at the end of this reaction?
Chemistry
1 answer:
juin [17]3 years ago
8 0
<span>2Al + 3Br2 -------------> 2AlBr3

</span>3 g Al = 0.11 mol Al. 

<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>

<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>

<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g. 
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>


</span>
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