68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. grams of magnesium hydroxide will form from this reaction, and (compound) is the limiting reagent. Blank 1:
Blank 2:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

Now we convert the given masses of reactants to moles, using their respective molar masses:

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and Mg(NO₃)₂ is the limiting reagent.

Now we calculate how many Mg(OH)₂ are produced, using the moles of the limiting reagent:

Finally we convert Mg(OH)₂ moles to grams:

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the answer is b. dilution