Chapter 24: Problem 15
Write a balanced equation to show the reaction between \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) How many grams of \(\mathrm{CaH}_{2}\) are needed to produce \(26.4 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at \(20^{\circ} \mathrm{C}\) and \(746 \mathrm{mmHg}\) ?
Short Answer
Expert verified
23.0 g of \( \mathrm{CaH}_2 \) is needed.
Step by step solution
01
Write the Balanced Chemical Equation
The reaction between calcium hydride \( \mathrm{CaH}_2 \) and water \( \mathrm{H}_2\mathrm{O} \) produces calcium hydroxide \( \mathrm{Ca(OH)_2} \) and hydrogen gas \( \mathrm{H}_2 \). The balanced equation is: \[ \mathrm{CaH}_2 + 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ca(OH)_2} + 2 \mathrm{H}_2} \]
02
Convert Conditions to STP for Gas
Use the ideal gas law to convert the given conditions to standard temperature and pressure (STP). The formula is \( PV = nRT \). Convert the given conditions \( 746 \text{ mmHg} = 0.9803 \, \text{atm} \) and \( 20^\circ \text{C} = 293\, \text{K} \). Thus, \[ n = \frac{PV}{RT} = \frac{(0.9803\, \text{atm})(26.4\, \text{L})}{(0.0821\, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol})(293\, \text{K})} \approx 1.095\, \text{mol} \]
03
Use Stoichiometry to Determine Moles of \(\mathrm{CaH}_2\)
From the balanced equation, 1 mole of \( \mathrm{CaH}_2 \) produces 2 moles of \( \mathrm{H}_2 \). Thus, to produce 1.095 moles of \( \mathrm{H}_2 \), we need \[ \frac{1.095}{2} = 0.5475\, \text{mol} \, \mathrm{CaH}_2 \]
04
Calculate Mass of \( \mathrm{CaH}_2 \) Needed
Calculate the mass of \( \mathrm{CaH}_2 \) using its molar mass \( \approx 42.09\, \text{g/mol} \). \[ \text{Mass of } \mathrm{CaH}_2 = 0.5475\, \text{mol} \times 42.09\, \text{g/mol} = 23.048\, \text{g} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation used in chemistry to describe the behavior of gases. It is expressed as \( PV = nRT \), where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant (approximately 0.0821 L atm / K mol),
- \( T \) is the temperature in Kelvin.
Stoichiometry
Stoichiometry is the part of chemistry that relates quantities of reactants and products in a chemical reaction. It uses the coefficients from a balanced chemical equation to determine the proportions of each substance involved. For example, in the reaction between calcium hydride \( \mathrm{CaH}_2 \) and water \( \mathrm{H}_2\mathrm{O} \):\[ \mathrm{CaH}_2 + 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Ca(OH)_2} + 2 \mathrm{H}_2} \]According to this equation, 1 mole of \( \mathrm{CaH}_2 \) produces 2 moles of \( \mathrm{H}_2 \). Thus, applying stoichiometry allows us to convert the moles of \( \mathrm{H}_2 \) (calculated from the ideal gas law) into the equivalent moles of \( \mathrm{CaH}_2 \) needed.
Molar Mass
Molar mass is a crucial concept in chemistry, referring to the mass of one mole of a substance. It is typically given in grams per mole (g/mol) and can be derived from the atomic masses of the elements that comprise the compound. For calcium hydride \( \mathrm{CaH}_2 \), the molar mass is calculated as:
- Calcium (Ca): Approximately 40.08 g/mol,
- Hydrogen (H): Approximately 1.01 g/mol \( \times 2 \) (due to two hydrogen atoms).
- Total for \( \mathrm{CaH}_2 \): 40.08 + (2 \times 1.01) \approx 42.09 \text{ g/mol}.
Gas Conditions
Gas conditions refer to the environment under which a gas is measured, including temperature, pressure, and volume. These conditions are essential in calculations involving gases since they directly affect gas behavior as described by gas laws.
- Temperature is typically measured in Celsius but converted to Kelvin for calculations involving the ideal gas law. Add 273.15 to the Celsius temperature to convert it to Kelvin.
- Pressure can be provided in various units, such as mmHg or atm. For ideal gas law applications, you often convert all pressure measurements to atmospheres (1 atm = 760 mmHg). In this exercise, 746 mmHg was converted to approximately 0.9803 atm.
- Volume is usually given in liters.