Question

How many litres of hydrogen gas can be generated by reacting \(6.25 \mathrm{~g}\) of barium hydride with water at \(20^{\circ} \mathrm{C}\) and \(755 \mathrm{~mm} \mathrm{Hg}\) pressure according to the following chemical equation: \(\mathrm{BaH}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})+2 \mathrm{H}_{2}(\mathrm{~g})\) (a) \(2.17\) litre (b) \(3.58\) litre (c) \(1.07\) litre (d) \(4.57\) litre

Short answer

The answer is (a) 2.17 litres.

Step-by-step solution

Calculate Moles of Barium Hydride

First, we need to calculate the number of moles of barium hydride (\(\text{BaH}_2\)) from its given mass. The molar mass of \(\text{BaH}_2\) is \(137.33 + 2 \times 1.008 = 139.346 \; \text{g/mol}\). Use the formula: \[ \text{moles of } \text{BaH}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{6.25 \; \text{g}}{139.346 \; \text{g/mol}} \approx 0.0448 \; \text{mol} \]

Calculate Moles of Hydrogen Gas Produced

From the balanced chemical equation:\[ \text{BaH}_2 + 2\text{H}_2\text{O} \rightarrow \text{Ba(OH)}_2 + 2\text{H}_2 \]Each mole of \(\text{BaH}_2\) produces 2 moles of \(\text{H}_2\). Therefore, \[ \text{moles of } \text{H}_2 = 2 \times 0.0448 \; \text{mol} = 0.0896 \; \text{mol} \]

Use Ideal Gas Law to Find Volume of Hydrogen

The ideal gas law relates pressure, volume, temperature, and moles via:\[ PV = nRT \]Where:- \( P \) is pressure (755 mmHg = 755/760 atm)- \( V \) is volume- \( n = 0.0896 \; \text{mol} \)- \( R = 0.0821 \; \text{L atm/mol K} \)- \( T = 20°C = 293 \; \text{K} \) Substitute: \[ (\frac{755}{760} \; \text{atm}) V = (0.0896 \; \text{mol})(0.0821 \; \text{L atm/mol K})(293 \; \text{K}) \]Solve for \( V \):\[ V \approx \frac{0.0896 \times 0.0821 \times 293}{0.9934} \approx 2.17 \; \text{L} \]

Match Volume with Given Options

The calculated volume of hydrogen gas is approximately \(2.17\) liters. Check which option matches this result: (a) \(2.17\) litre (b) \(3.58\) litre (c) \(1.07\) litre (d) \(4.57\) litre Therefore, option (a) is correct.

Key concepts

Chemical Reactions

In chemistry, reactions are at the core of understanding how substances interact. A chemical reaction involves the transformation of reactants into products. For this exercise, we have the chemical reaction: \( \mathrm{BaH}_2(\mathrm{~s}) + 2 \mathrm{H}_2 \mathrm{O}(1) \rightarrow \mathrm{Ba}(\mathrm{OH})_2(\mathrm{aq}) + 2 \mathrm{H}_2(\mathrm{~g}) \). This equation shows us that barium hydride reacts with water to produce barium hydroxide and hydrogen gas.

To balance the equation, we must ensure that the number of atoms for each element is the same on both sides. In this balanced equation, one mole of \( \text{BaH}_2 \) produces two moles of hydrogen gas. Recognizing this stoichiometric relationship is crucial for determining how much hydrogen gas can be generated from a given amount of barium hydride.

Molar Mass Calculation

Calculating the molar mass is essential for converting between mass and moles, which are more convenient for stoichiometric calculations. The molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol).

In our problem, we needed to find the molar mass of barium hydride, \( \text{BaH}_2 \), to convert the given mass into moles. Here's how it's calculated:

• Barium (Ba) has an atomic mass of approximately 137.33 g/mol.
• Hydrogen (H) has an atomic mass of approximately 1.008 g/mol.

Using these, the molar mass of \( \text{BaH}_2 \) is calculated as follows:
\[ 137.33 + 2 \times 1.008 = 139.346 \, \text{g/mol} \]
This conversion allows us to calculate the moles of \( \text{BaH}_2 \) present in a sample, which is essential for determining how much hydrogen gas can be produced.

Ideal Gas Law

The ideal gas law is a useful equation that connects the state variables of a gas: pressure (P), volume (V), temperature (T), and the amount in moles (n). The equation is represented as \( PV = nRT \), where \( R \) is the ideal gas constant. This relationship is key in predicting the behavior of gases under different conditions.

To use the ideal gas law in calculations, we:

• Convert pressure into atmospheres (since \( R = 0.0821 \, \text{L atm/mol K} \)). For example, \( 755 \text{ mmHg} \) is converted by dividing by 760, resulting in \( \approx 0.9934 \text{ atm} \).
• Convert temperature from Celsius to Kelvin by adding 273 (e.g., \( 20^{\circ} \text{C} \to 293 \, \text{K} \)).
• Use the given moles of gas, plug these values into the formula, and solve for volume \( V \).

Using the ideal gas law in our situation helped relate the conditions of temperature and pressure to find how many liters of hydrogen gas can be generated.

Pressure Conversion

Converting pressure units is fundamental when applying the ideal gas law. Pressure is often measured in different units like atmospheres (atm), millimeters of mercury (mmHg), or torr. These variations need to be unified to be consistent with the gas constant \( R \), which is usually \( 0.0821 \, \text{L atm/mol K} \).

In this exercise, the original pressure was given in mmHg: \( 755 \text{ mmHg} \). To incorporate it into the ideal gas law, we converted it to atmospheres:

• Use the conversion factor where \( 1 \text{ atm} = 760 \text{ mmHg} \).
• Apply the conversion: \( 755 \text{ mmHg} \div 760 \text{ mmHg/atm} \approx 0.9934 \text{ atm} \).

This conversion ensures that all measurements are in compatible units within the formula, providing precise calculation of the gas behavior under given conditions.