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If you start with 10.0 grams of lithium hydroxide and an excess of hydrogen bromide, how many grams of lithium bromide will be produced? A. 2.76 g B. 36.3 g C. 86.8 g D. 210 g
If you start with 10.0 grams of lithium hydroxide and an excess of hydrogen bromide, 36.3 of lithium bromide will be produced. ***LiOH + HBr - -> LiBr + H2O (? = coefficient in the balanced equation) mass of x = (mole of y) * (? mol x / ? mol y) [ * (molar mass of x) mass of LiBr = (10 g / 23.95 g/mol) * (1 mol LiBr / 1 mol LiOH) * (86.85 g/mol LiBr) mass of LiBr = 36.3 g 36.3 grams ]
Expert answered|emdjay23|Points 1766|
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Asked 7/2/2013 12:15:35 PM
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