MgO(s)+H2O(l)->Mg(OH)2(aq)
You want 98.9 g Mg(OH)2. How many mols is that? mols = g/molar mass = approx 1.5 but you need to recalculate all of these estimates to get a better figure.
Convert mols Mg(OH)2 to mols Mg) using the coefficients in the balanced equation above. That is a 1:1 ratio; therefore mols MgO = approx 1.5.
Now convert mols MgO to grams MgO. That is mols MgO x molar mass MgO = approx 70g. That gives you the mass MgO needed in a 100% yield problem. It is is only 81% then you need about 70/0.81 = ? g MgO.
Post your work if you get stuck.
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?
MgO(s)+H2O(l)->Mg(OH)2(aq)
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