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Answer 1 · 2017-11-22T22:27:09

See below.

We already have a balanced chemical equation, so now the only thing we have to do is the stoichiometry:

Step 1: Convert grams of $L i O H$ $LiOH$ into moles using its molar mass.

$10.0 g LiOH \cdot (\frac{1 m o l L i O H}{23.948 g r a m s}) = 0.418 mols LiOH$ $"10.0 g LiOH"*((1 mol LiOH)/(23.948 grams))="0.418 mols LiOH"$

Step 2: Use the mol ratio of $L i O H$ $LiOH$ to $L i B r$ $LiBr$ to find how many mols of $L i B r$ $LiBr$ will be produced.

$0.418 mols LiOH \cdot (\frac{1 m o l L i B r}{1 m o l L i O H}) = 0.418 mols LiBr$ $"0.418 mols LiOH"*((1 mol LiBr)/(1 mol LiOH))="0.418 mols LiBr"$

Step 3: Convert mols of $L i B r$ $LiBr$ to grams using its molar mass.

$0.418 mols LiBr \cdot (\frac{86.845 g}{1 m o l L i B r}) = 36.3 grams LiBr$ $"0.418 mols LiBr"*((86.845g)/(1 mol LiBr))="36.3 grams LiBr"$

Hope this helps!