Answer :
Final answer:
The concentration of 2L LiOH required to neutralize 1L of 1.5M H₂CO₃ is 1.5M, as determined from the stoichiometry of the balanced chemical equation.
Explanation:
To determine the concentration of LiOH needed to neutralize 1L of 1.5M H₂CO₃ (carbonic acid), we first need to write the balanced chemical equation for the reaction. LiOH will react with H₂CO₃ to form Li₂CO₃ and water (H₂O), as shown below:
2 LiOH + H₂CO₃ → Li₂CO₃ + 2 H₂O
From the balanced equation, we can see that it takes 2 moles of LiOH to neutralize 1 mole of H₂CO₃. Given that we have 1L of 1.5M H₂CO₃, we can find the number of moles of H₂CO₃:
moles H₂CO₃ = (1.5M) × (1L) = 1.5 moles
Since the reaction ratio is 2:1, we need 2 moles of LiOH for every mole of H₂CO₃:
moles LiOH = 2 × 1.5 moles = 3 moles
To find the concentration of LiOH, we divide the moles of LiOH by the volume of the LiOH solution:
concentration LiOH = 3 moles / 2L = 1.5M
Therefore, the concentration of 2L LiOH needed to neutralize 1L of 1.5M H₂CO₃ is 1.5M.
Final answer:
To neutralize 1L of 1.5M H2CO2, a concentration of 0.75M LiOH is needed.
Explanation:
To find the concentration of LiOH, we can use the formula C1V1 = C2V2. Given that the volume (V) of H2CO2 is 1L and its concentration (C) is 1.5M, we can substitute these values into the formula: C1(2L) = (1.5M)(1L). Solving for C1, we get C1 = (1.5M)(1L)/(2L) = 0.75M.
Therefore, the concentration of 2L LiOH needed to neutralize 1L of 1.5M H2CO2 is 0.75M. Therefore, option a) 0.5M is the correct answer.
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