For the neutralization reaction involving HNO₃ and Ca(OH)₂, liters of 1.55 M HNO₃ are needed to react with 45.8 ml of a 4.66 m Ca(OH)₂ solution is 0.270 L.
<h3>What is neutralization reaction?</h3>
Those reactions in which acids and bases will combine with each other for the formation of water molecule and salt, is known as neutralization reaction.
Given chemical reaction is:
2HNO₃ + Ca(OH)₂ → 2H₂O + Ca(NO₃)₂
Moles (n) of Ca(OH)₂ will be calculated by using the below formula:
M = n/V, where
M = molarity = 4.66M
V = volume = 45.8 mL = 0.045 L
n = (4.66)(0.045) = 0.2097 mol
From the stoichiometry of the reaction,
1 mole of Ca(OH)₂ = reacts with 2 moles of HNO₃
0.2097 mole of Ca(OH)₂ = reacts with 2×0.2097=0.419 moles of HNO₃
Again by using the molarity equation volume will be calculated for HNO₃ as:
V = (0.419) / (1.55) = 0.270 L
Hence required volume of HNO₃ is 0.270 L.
To know more about neutralization reaction, visit the below link:
brainly.com/question/23008798
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