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17. How much 4M Mg(OH)₂ do you need to neutralize 8L of 2M H₂SO4?

Given: CV-C₂V₂
a)4L b)6L c)8L d)16L e)32L

User Maanijou
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1 Answer

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Final answer:

To neutralize 8L of 2M H₂SO4, you will need 2L of 4M Mg(OH)₂.


Step-by-step explanation:

To find the amount of 4M Mg(OH)₂ needed to neutralize 8L of 2M H₂SO4, you can use the formula CV = C₂V₂. Here, C represents the concentration and V represents the volume. First, determine the moles of H₂SO4 using the formula moles = concentration × volume. So, moles of H₂SO4 = 2M × 8L = 16 moles. Since the ratio between Mg(OH)₂ and H₂SO4 is 2:1, you will need half the amount of moles of Mg(OH)₂. So, moles of Mg(OH)₂ = 16/2 = 8 moles. Now, use the formula V₂ = moles ÷ concentration to find the volume of 4M Mg(OH)₂. Here, moles = 8 and concentration = 4M. So, V₂ = 8 moles ÷ 4M = 2L.

Therefore, you will need 2L of 4M Mg(OH)₂ to neutralize 8L of 2M H₂SO4.


Learn more about Neutralization of acids and bases

User Andrew Sasha
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