The neutralization reaction between hydrochloric acid and calcium hydroxide produces calcium chloride and water. 2 HCl + Ca(OH)2 = 2 H2O + CaCl2 If 0.75 grams of Ca(OH)2 is used in the reaction, determine the following: a. Number of moles of HCl(MM=36.46 g/mol) needed to completely react with Ca(OH)2. The molar mass of Ca(OH)2 is 74.10 g/mol. b. Mass of CaCl2(MM=110.98 g/mol) formed in the reaction.
The neutralization reaction between hydrochloric acid and calcium hydroxide produces calcium chloride and water.
2 HCl + Ca(OH)2 = 2 H2O + CaCl2
If 0.75 grams of Ca(OH)2 is used in the reaction, determine the following:
a. Number of moles of HCl(MM=36.46 g/mol) needed to completely react with Ca(OH)2. The molar mass of Ca(OH)2 is 74.10 g/mol.
b. Mass of CaCl2(MM=110.98 g/mol) formed in the reaction.
The Correct Answer and Explanation is :
To solve the problem, let’s address each part step-by-step.
Given Data:
- Reaction: ( 2 \, \text{HCl} + \text{Ca(OH)}_2 \rightarrow 2 \, \text{H}_2\text{O} + \text{CaCl}_2 )
- Mass of ( \text{Ca(OH)}_2 ): 0.75 g
- Molar mass of ( \text{Ca(OH)}_2 ): 74.10 g/mol
- Molar mass of ( \text{HCl} ): 36.46 g/mol
- Molar mass of ( \text{CaCl}_2 ): 110.98 g/mol
a. Moles of ( \text{HCl} ) Required:
- Determine the moles of ( \text{Ca(OH)}_2 ):
[
\text{Moles of } \text{Ca(OH)}_2 = \frac{\text{Mass of } \text{Ca(OH)}_2}{\text{Molar mass of } \text{Ca(OH)}_2}
]
[
\text{Moles of } \text{Ca(OH)}_2 = \frac{0.75 \, \text{g}}{74.10 \, \text{g/mol}} = 0.01012 \, \text{mol}
] - Use stoichiometry to calculate moles of ( \text{HCl} ):
The balanced equation shows that 1 mole of ( \text{Ca(OH)}_2 ) reacts with 2 moles of ( \text{HCl} ). Thus:
[
\text{Moles of } \text{HCl} = 0.01012 \, \text{mol} \times 2 = 0.02024 \, \text{mol}
]
b. Mass of ( \text{CaCl}_2 ) Formed:
- Use stoichiometry to calculate moles of ( \text{CaCl}_2 ):
From the balanced equation, 1 mole of ( \text{Ca(OH)}_2 ) produces 1 mole of ( \text{CaCl}_2 ). Therefore:
[
\text{Moles of } \text{CaCl}_2 = 0.01012 \, \text{mol}
] - Calculate the mass of ( \text{CaCl}_2 ):
[
\text{Mass of } \text{CaCl}_2 = \text{Moles of } \text{CaCl}_2 \times \text{Molar mass of } \text{CaCl}_2
]
[
\text{Mass of } \text{CaCl}_2 = 0.01012 \, \text{mol} \times 110.98 \, \text{g/mol} = 1.123 \, \text{g}
]
Final Answers:
a. Moles of ( \text{HCl} ) needed: ( 0.02024 \, \text{mol} )
b. Mass of ( \text{CaCl}_2 ) formed: ( 1.123 \, \text{g} )
Explanation:
The neutralization reaction is a double displacement reaction between hydrochloric acid and calcium hydroxide. Based on the molar ratio from the balanced chemical equation, ( \text{Ca(OH)}_2 ) reacts with twice the moles of ( \text{HCl} ), and the amount of ( \text{CaCl}_2 ) produced is directly proportional to the moles of ( \text{Ca(OH)}_2 ). Using the given masses and molar masses, we converted grams to moles, used stoichiometric relationships, and calculated the required quantities step-by-step. This method ensures accurate results consistent with the law of conservation of mass.