well, you could (at perfect yield), gotten 98.0/.81=121grams, or moles Mg(OH)2 121/58.3=2.07 moles
MgO+H2O>>>Mg(OH)2. From the balanced equation, you needed then 2.07 moles MgO, or 2.07*formulamassMgO grams
Magnesium hydroxide is produced by reacting magnesium oxide with excess water. If the reaction has an expected yield of 81%, how much magnesium oxide should be reacted to produce 98.0 g of magnesium hydroxide?
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