Question

What weight of \(\mathrm{Ca}(\mathrm{OH})_{2}\), calcium hydroxide is needed to neutralize \(28.0 \mathrm{~g}\) of \(\mathrm{HCl}\). What weight of phosphoric acid would this weight of \(\mathrm{Ca}(\mathrm{OH})_{2}\) neutralize?

Short answer

Approximately 28.4 g of calcium hydroxide is needed to neutralize 28 g of hydrochloric acid, and the same amount of calcium hydroxide (28.4 g) would neutralize 25.1 g of phosphoric acid.

Step-by-step solution

Convert the mass of \( \mathrm{HCl} \) to moles

To convert the mass of \( \mathrm{HCl} \) to moles, we need to use the molar mass of \( \mathrm{HCl} \), which is \(36.5 \, \mathrm{g/mol}\). The formula to convert mass to moles is \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \] Plugging in the given values, we get \[ \text{Moles of } \mathrm{HCl} = \frac{28.0 \, \mathrm{g}}{36.5 \, \mathrm{g/mol}} \approx 0.767 \, \mathrm{mol} \].

Determine the moles of \( \mathrm{Ca(OH)}_{2} \) needed

According to the balanced equation, one mole of \( \mathrm{Ca(OH)}_{2} \) reacts with two moles of \( \mathrm{HCl} \). So, we need half as many moles of \( \mathrm{Ca(OH)}_{2} \) as \( \mathrm{HCl} \). Hence, moles of \( \mathrm{Ca(OH)}_{2} \) needed = \( \frac{0.767 \, \mathrm{mol}}{2} = 0.384 \, \mathrm{mol} \).

Calculate the mass of \( \mathrm{Ca(OH)}_{2} \)

We already know the number of moles of \( \mathrm{Ca(OH)}_{2} \) needed. Now, we will convert this back to mass using its molar mass, which is \(74.1 \, \mathrm{g/mol}\). The formula to convert moles to grams is \[ \text{Mass} = \text{Moles} \times \text{Molar mass} \]. Implementing the values we have, we get \[ \text{Mass of } \mathrm{Ca(OH)}_{2} = 0.384 \, \mathrm{mol} \times 74.1 \, \mathrm{g/mol} \approx 28.4 \, \mathrm{g} \].

Determine the moles of \( \mathrm{H}_{3}\mathrm{PO}_{4} \) neutralized by the same mass of \( \mathrm{Ca(OH)}_{2} \)

According to the balanced equation, three moles of \( \mathrm{Ca(OH)}_{2} \) react with two moles of \( \mathrm{H}_{3}\mathrm{PO}_{4} \). We have \( \mathrm{Ca(OH)}_{2} \) in the quantity of \(0.384 \, \mathrm{mol}\), so the moles of \( \mathrm{H}_{3}\mathrm{PO}_{4} \) neutralized will be \[ \frac{2}{3} \times 0.384 \, \mathrm{mol} \approx 0.256 \, \mathrm{mol} \].

Calculate the weight of neutralized \( \mathrm{H}_{3}\mathrm{PO}_{4} \)

We have the moles of neutralized \( \mathrm{H}_{3}\mathrm{PO}_{4} \), and we know its molar mass (\(98.0 \, \mathrm{g/mol}\)). By using the formula \[ \text{Mass} = \text{Moles} \times \text{Molar mass} \] we can find the mass of neutralized \( \mathrm{H}_{3}\mathrm{PO}_{4} \) as follows: \[ \text{Mass of } \mathrm{H}_{3}\mathrm{PO}_{4} = 0.256 \, \mathrm{mol} \times 98.0 \, \mathrm{g/mol} \approx 25.1 \, \mathrm{g} \]. In summary, it would take approx '28.4 g' of calcium hydroxide to neutralize '28 g' of hydrochloric acid and the same amount of calcium hydroxide ('28.4 g') would neutralize '25.1 g' of phosphoric acid.

Key concepts

Calcium Hydroxide

Calcium hydroxide, commonly referred to as slaked lime, is a chemical compound with the formula \( \mathrm{Ca(OH)}_{2} \). It appears as a white powder and is slightly soluble in water, forming a solution known as lime water. This compound finds various applications, such as in the field of construction (as a mortar ingredient), water purification, and most importantly, in chemical reactions.

In the context of stoichiometry problems, it serves as a base that can neutralize acids. Its molar mass is \( 74.1 \, \mathrm{g/mol} \), which is essential information for calculating its participation in chemical reactions. By understanding its properties and applications, students can appreciate its role in stoichiometry-related calculations.

Neutralization Reaction

A neutralization reaction is a chemical reaction in which an acid and a base react quantitatively with each other. In a reaction in aqueous solution, the acid and base neutralize each other, forming water and a salt. It is a vital concept in chemistry, especially in stoichiometry where balanced equations are used to calculate reactants and products.

For instance, when calcium hydroxide (\(\mathrm{Ca(OH)}_{2}\)) reacts with hydrochloric acid (\(\mathrm{HCl}\)), the result is the formation of water and calcium chloride, as shown in the balanced equation below:

• \( \mathrm{Ca(OH)}_{2} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_{2} + 2\mathrm{H}_{2} \mathrm{O} \)

Understanding these reactions is crucial for solving problems involving the determination of the amount of base needed to neutralize a given amount of acid.

Mole Concept

The mole concept is a fundamental principle in chemistry that allows chemists to count particles by weighing them. One mole is usually defined as \( 6.022 \times 10^{23} \) particles (Avogadro’s number) and is equivalent to the substance’s atomic or molecular mass expressed in grams.

In stoichiometry, the mole concept helps convert between mass and moles using the substance's molar mass. For example, to find out how many moles of \(\mathrm{HCl}\) correspond to a given mass, we use its molar mass \( (36.5 \, \mathrm{g/mol}) \) with the formula:

• \( \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \)

This concept is key in performing stoichiometric calculations to determine the reactants and products involved in a chemical transfer.

Chemical Equations

Chemical equations represent the reactants and products in a chemical reaction using symbols and formulas. These equations are balanced to obey the law of conservation of mass, meaning the number of atoms of each element on the reactants' side must equal the number on the products' side.

Balancing a chemical equation involves ensuring the quantities of substances in moles match across the reaction. For example, the balanced equation for the neutralization of \(\mathrm{HCl}\) with \(\mathrm{Ca(OH)}_{2}\) demonstrates that one mole of \(\mathrm{Ca(OH)}_{2}\) reacts with two moles of \(\mathrm{HCl}\).

Similarly, for the neutralization with phosphoric acid \((\mathrm{H}_{3}\mathrm{PO}_{4})\), the balanced equations guide the stoichiometric calculations, determining how many moles of one reactant are needed to react completely with another. This balance is integral for converting between masses and moles, allowing for precise solutions in chemical analysis.