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.
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.
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