Problem 58 - Chemistry

Question

A quantity of \(0.225 \mathrm{~g}\) of a metal \(\mathrm{M}\) (molar mass = \(27.0 \mathrm{~g} / \mathrm{mol}\) ) liberated \(0.303 \mathrm{~L}\) of molecular hydrogen (measured at \(17^{\circ} \mathrm{C}\) and \(741 \mathrm{mmHg}\) ) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of M.

Step-by-Step Solution

Answer

The balanced equation is \(2M + 6HCl \rightarrow 3H2 + 2MCl3\). The formula for the oxide of M is \(M2O3\) and for the sulfate of M is \(M2(SO4)3\).

Step 1: Calculate Number of Moles of Hydrogen (H2)

Convert volume of H2 to moles using Avogadro's law, which states that one mole of any gas occupies a volume of \(22.4 L\) at STP. The STP conditions are \(0^{\circ}C\) (which is \(273.15 K\)) and \(1 atm\), but in this case we have different conditions; \(17^{\circ}C\) (which is \(290.15 K\)) and \(741 mmHg\) (which is approximately \(0.97 atm\)). The equation to use is \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant and \(T\) is the temperature in Kelvin. Rearranging for n gives \(n = PV/RT\). After substituting given values, we find number of moles: \(n_{H2} = (0.97*0.303)/(0.0821*290.15) = 0.012 moles.\)

Step 2: Calculate Number of Moles of Metal M

Now, calculate the number of moles of metal \(M\). The number of moles (\(n\)) is given by mass divided by molar mass. Using the given mass and molar mass: \(n_M = 0.225/27 = 0.00833 moles\)

Step 3: Deduce the Stoichiometry of the Reaction

Comparing the moles of \(M\) to the moles of \(H2\), one has a ratio of 0.00833 to 0.012, which simplifies to 2:3. This means two moles of \(M\) react to produce three moles of \(H2\). This information is used to write the balanced equation: \(2M + 6HCl \rightarrow 3H2 + 2MCl3\)

Step 4: Write Formulas for the Oxide and Sulfate of M

Considering that \(M\) forms a chloride \(MCl3\) as per the reaction, this means the metal is trivalent, i.e., \(M^{3+}\). Thus, the formula for the oxide of \(M\) must be \(M2O3\) (since the oxide ion \(O^{2-}\) requires three \(M^{3+}\) ions for charge balance), and for the sulfate must be \(M2(SO4)3\) (since the sulfate ion \(SO4^{2-}\) requires three \(M^{3+}\) ions for charge balance).

Key Concepts

Chemical reactions

Chemical reactions involve the transformation of one or more substances into different substances. This change is expressed through a balanced chemical equation. Each side of the equation represents either the reactants or the products. In our example exercise, we have a reaction where metal \( M \) reacts with hydrochloric acid (HCl) to produce molecular hydrogen \( H_2 \) and \( MCl_3 \), the chloride of metal \( M \). The process involves the rearrangement of atoms and the breaking and forming of bonds.

By analyzing the moles of metal \( M \) and hydrogen \( H_2 \) produced, we determine the stoichiometry of our reaction, represented by the balanced equation. This stoichiometry uses the mole ratio to ensure the conservation of mass, which states that the mass of the reactants must equal the mass of the products.

• The given exercise leads to the balanced chemical equation: \( 2M + 6HCl \rightarrow 3H_2 + 2MCl_3 \).
• This equation indicates how two moles of metal \( M \) react with six moles of hydrochloric acid to produce three moles of molecular hydrogen and two moles of \( MCl_3 \).

Understanding such reactions is crucial for analyzing the chemical behavior of substances.

Molar mass

Molar mass is a fundamental concept in chemistry that helps convert between the mass of a substance and the number of moles. It is expressed in grams per mole (g/mol) and represents the mass of one mole of a substance. This concept is particularly useful in calculating the amount of a substance needed or produced in a chemical reaction.

For the exercise, we were provided the molar mass of metal \( M \) as 27 g/mol. Using the formula:

• \( n = \frac{\text{mass}}{\text{molar mass}} \)
• where \( n \) is the number of moles, the mass of the metal given was 0.225 g. By substituting into the formula, we calculate the moles of \( M \): \( n_M = \frac{0.225}{27} = 0.00833 \text{ moles} \).

This calculation is vital for determining how the moles of one reactant relate to the moles of other reactants or products in the balanced chemical equation.

Gas laws

Gas laws describe how gases behave and relate properties such as pressure, volume, and temperature. In the context of our exercise, understanding the behavior of gases is essential for converting the given volume of \( H_2 \) gas to moles under non-standard conditions, using the ideal gas law equation:

• \( PV = nRT \)
• where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.

In our example, the external conditions were a temperature of 17°C (which is 290.15 K) and a pressure of 741 mmHg (approximately 0.97 atm). These values allowed us to solve for \( n \), the moles of \( H_2 \):

\[ n_{H2} = \frac{0.97 \cdot 0.303}{0.0821 \cdot 290.15} = 0.012 \text{ moles} \]

This demonstrates how the gas laws facilitate the conversion of volume to moles, which is crucial for stoichiometric calculations.

Chemical formulas

Chemical formulas are shorthand representations of the elemental composition of molecules and compounds. They provide essential information about the number and type of atoms that make up a substance. For elements like metal \( M \), identifying the corresponding chemical formulas for its compounds is significant.

From the reaction, \( M \) forms \( MCl_3 \), indicating it is trivalent, meaning it forms ions with a charge of 3+. Using this information, we can deduce the formulas for other compounds of \( M \).

• The oxide of \( M \) would be \( M_2O_3 \) as \( M^{3+} \) requires two \( O^{2-} \) ions for charge balance.
• Similarly, the sulfate of \( M \) would be \( M_2(SO_4)_3 \) because each \( SO_4^{2-} \) ion requires three \( M^{3+} \) ions to achieve neutrality.

These formulas are tools chemists use to represent the composition of compounds accurately and succinctly.