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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.
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Answer:

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:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

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

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

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:

  • 0.528 mol Mg(NO₃)₂ * [tex]\frac{1molMg(OH)_2}{1molMg(NO_3)_2}[/tex] = 0.528 mol Mg(OH)₂

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

  • 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g

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