Aluminum and Bromine reaction?

When 2.70 g of Al reacts with excess Br2, the maximum mass of Al2Br6 that can be produced is?

2Al + 3Br ---> Al2Br6

When you calculate the number of moles of Al taking place in the reaction... is it

2.70/27 = 0.1 moles?


2.70/54 = 0.05 moles?

I am going with the first 1...

0.1 moles of Al produces 0.05 moles of Al2Br6

so 0.05 x ((2 x 27)+(6 x 79.9)) = 26.67 g of Al2Br6


2 Answers

  • 8 years ago
    Favorite Answer

    n = 2.7/27 = 0.1 moles. That's how you start the problem after you've balanced your equation.

    The rest of what you did is correct.

  • 8 years ago

    Yes, the first one is correct.

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