At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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.
What is neutralization reaction?
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:
https://brainly.com/question/23008798
#SPJ4
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.