Video: Calculating the Amount of Mg(OH)₂ Required to Neutralize a Given Mass of HNO₃, Given the Molar Mass of HNO₃

How many moles of Mg(OH)₂ are required to neutralize 126 g of HNO₃? (Molar mass of HNO₃ = 63 g/mole) [A] 0.25 moles [B] 0.50 moles [C] 1.0 moles [D] 2.0 moles [E] 4.0 moles

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Video Transcript

How many moles of Mg(OH)2 are required to neutralize 126 grams of HNO3? Molar mass of HNO3 is equal to 63 grams per mole. (A) 0.25 moles, (B) 0.50 moles, (C) 1.0 moles, (D) 2.0 moles, (E) 4.0 moles.

In this problem, we’re reacting Mg(OH)2 or magnesium hydroxide, which is a strong base, with HNO3 or nitric acid, which is a strong acid. When an acid and a base react, they neutralize each other to form a salt and water. So in this reaction, we’ll have nitric acid reacting with magnesium hydroxide to form magnesium nitrate and water.

Before we do anything else, let’s balance this equation. Since the polyatomic anion nitrate sticks together over the course of this reaction and doesn’t break up into its constituent elements, we’ll balance it as one group. So we have one nitrate on the reactant side and two nitrates on the product side. We have one magnesium on each side of the reaction, so we’re good to go there. We have four hydrogens on the reactant side, but only two on the product side. So we’ll put a two in front of the water. And our last element is oxygen, which we have two of on both sides of the reaction. So we’re all balanced.

The question tells us that we have 126 grams of nitric acid to start off with. And we want to know how many moles of magnesium hydroxide we need to neutralize or completely react with this amount of nitric acid. To solve this problem, we’ll first want to convert the mass of nitric acid that we have into the amount of nitric acid we have in moles. From there, we’ll be able to solve the problem and find the amount of magnesium hydroxide in moles by using the balanced stoichiometric coefficients from a reaction.

So let’s tackle the first step of this problem. We need to convert 126 grams of HNO3 into moles of HNO3. We can do this by using the molar mass of HNO3 that was given to us in the problem, 63 grams per mole. If we divide the mass by the molar mass, we’ll get the amount of HNO3 in moles. 126 divided by 63 gives us two. So we have two moles of nitric acid.

Now, let’s do the second step, converting from moles of nitric acid to moles of magnesium hydroxide. We’ll need the stoichiometric coefficients from our balanced chemical equation. As we can see for every two moles of nitric acid, we need one mole of magnesium hydroxide. So using this ratio, we can see that to neutralize our two moles of nitric acid, we would need one mole of magnesium hydroxide, which matches answer choice (C).

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