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Given the balanced equation

2Al + 6HBr --> 2AlBr3 + 3H2

In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).

150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr

----------- 266.6g AlBr3 * 2 molecules AlBr3

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