Question

For each of the following tunbalanced equations, calculate how many moles of the second reactant would be required to react completely with 0.557 grams of the first reactant. a. \(\mathrm{Al}(s)+\mathrm{Br}_{2}(l) \rightarrow \mathrm{AlBr}_{3}(s)\) b. \(\mathrm{Hg}(s)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Hg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{K}(s)+\mathrm{P}(s) \rightarrow \mathrm{K}_{3} \mathrm{P}(s)\) d. \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\)

Short answer

a. 0.0309 moles of Br2 are required. b. 0.00557 moles of HClO4 are required. c. 0.00474 moles of P are required. d. 0.139 moles of Cl2 are required.

Step-by-step solution

Balance the chemical equation

The balanced equation is: $$2\mathrm{Al}(s)+3\mathrm{Br}_{2}(l) \rightarrow 2\mathrm{AlBr}_{3}(s)$$

Convert grams to moles

Use the molar mass of Al (26.98 g/mol) to convert the given mass of Al to moles: $$0.557\,\text{g Al} \times \frac{1\,\text{mol Al}}{26.98\,\text{g Al}} = 0.0206\,\text{mol}\, \mathrm{Al}$$

Use stoichiometry to find moles of second reactant required

From the balanced equation, 2 mol Al reacts with 3 mol Br2: $$0.0206\,\text{mol Al} \times \frac{3\,\text{mol Br}_{2}}{2\,\text{mol Al}} = 0.0309\,\text{mol}\, \mathrm{Br}_{2}$$ Answer for a: 0.0309 moles of Br2 are required. b. Reaction: $\mathrm{Hg}(s)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Hg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)$

Balance the chemical equation

The balanced equation is: $$\mathrm{Hg}(s)+2\,\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Hg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)$$

Convert grams to moles

Use the molar mass of Hg (200.59 g/mol) to convert the given mass of Hg to moles: $$0.557\,\text{g Hg} \times \frac{1\,\text{mol Hg}}{200.59\,\text{g Hg}} = 0.00278\,\text{mol}\, \mathrm{Hg}$$

Use stoichiometry to find moles of second reactant required

From the balanced equation, 1 mol Hg reacts with 2 mol HClO4: $$0.00278\text{mol Hg} \times \frac{2\,\text{mol HClO}_{4}}{1\,\text{mol Hg}} = 0.00557\,\text{mol}\, \mathrm{HClO}_{4}$$ Answer for b: 0.00557 moles of HClO4 are required. c. Reaction: \(\mathrm{K}(s)+\mathrm{P}(s) \rightarrow \mathrm{K}_{3} \mathrm{P}(s)\)

Balance the chemical equation

The balanced equation is: $$3\,\mathrm{K}(s)+\mathrm{P}(s) \rightarrow \mathrm{K}_{3} \mathrm{P}(s)$$

Convert grams to moles

Use the molar mass of K (39.10 g/mol) to convert the given mass of K to moles: $$0.557\,\text{g K} \times \frac{1\,\text{mol K}}{39.10\,\text{g K}} = 0.0142\,\text{mol}\, \mathrm{K}$$

Use stoichiometry to find moles of second reactant required

From the balanced equation, 3 mol K reacts with 1 mol P: $$0.0142\,\text{mol K} \times \frac{1\,\text{mol P}}{3\,\text{mol K}} = 0.00474\,\text{mol}\, \mathrm{P}$$ Answer for c: 0.00474 moles of P are required. d. Reaction: $\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)$

Balance the chemical equation

The balanced equation is: $$\mathrm{CH}_{4}(g)+4\,\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+4\,\mathrm{HCl}(g)$$

Convert grams to moles

Use the molar mass of CH4 (16.04 g/mol) to convert the given mass of CH4 to moles: $$0.557\,\text{g CH}_{4} \times \frac{1\,\text{mol CH}_{4}}{16.04\,\text{g CH}_{4}} = 0.0347\,\text{mol}\, \mathrm{CH}_{4}$$

Use stoichiometry to find moles of second reactant required

From the balanced equation, 1 mol CH4 reacts with 4 mol Cl2: $$0.0347\,\text{mol CH}_{4} \times \frac{4\,\text{mol Cl}_{2}}{1\,\text{mol CH}_{4}} = 0.139\,\text{mol}\, \mathrm{Cl}_{2}$$ Answer for d: 0.139 moles of Cl2 are required.

Key concepts

Chemical Reactions

Understanding chemical reactions is crucial for mastering stoichiometry. A chemical reaction involves the transformation of reactants into products through the breaking and forming of chemical bonds. The key to solving stoichiometry problems is to start with a balanced chemical equation. This provides the necessary ratios, called stoichiometric coefficients, which denote how many moles of each reactant and product participate in the reaction.

For example, the balanced equation for the reaction between aluminum (Al) and bromine (Br₂) to form aluminum bromide (AlBr₃) is:
\[2\mathrm{Al}(s) + 3\mathrm{Br}_2(l) \rightarrow 2\mathrm{AlBr}_3(s)\].

This tells us that two moles of Al react with three moles of Br₂ to produce two moles of AlBr₃. Balancing equations helps us ensure the conservation of mass and atoms, which is a foundational principle in chemistry.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of that substance. It is a bridge between the atomic scale and the macroscopic scale, allowing chemists to count particles by weighing them. Calculating molar mass is essential for converting between grams of a substance and moles, and it's defined as the sum of the masses of all atoms in a molecule.

For instance, the molar mass of aluminum (Al) is 26.98 g/mol. This information allows us to construct a conversion factor:
\[1 \text{mol Al} = 26.98 \text{g Al}\].

Use this factor to convert a sample's mass to moles for stoichiometric calculations. To determine the moles of Al present in a 0.557-gram sample, divide the mass of the sample by the molar mass of Al, revealing the moles of Al in the sample.

Moles to Grams Conversion

Converting moles to grams is a fundamental aspect of stoichiometry and often required when you need to measure out chemicals for a reaction. Once you calculate the number of moles needed for a reaction, as determined by the balanced chemical equation, you often have to convert it back to grams to be able to weigh the substance.

The conversion is straightforward if you know the molar mass of the substance, which acts as a conversion factor. To convert moles to grams, multiply the number of moles by the molar mass of the substance in g/mol:
\[\text{grams} = \text{moles} \times \text{molar mass (g/mol)}\].

This will give you the mass in grams you'd theoretically need to react completely with the other reactant, adhering to the stoichiometric ratios from the chemical equation. By mastering this conversion process, students can seamlessly transition between the microscopic view of moles and the practical, measurable quantity of grams.