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3 years ago

How many milliliters of 0.25M H2SO4 can be prepared from 57 mL of a 3.0M solution of H2SO4?

Chemistry

1 answer:

DochEvi [55]3 years ago

5 0

Answer:

Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?

Explanation:

Now, we have 0.025 M H2SO4 and we do not know how much volume we have.

We will use the standard N1 X V1 = N2 X V2 for this calculation.

N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.

So V1= (0.03 X 525)/(0.025) = 630 ml.

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3 years ago

How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl2 according to the balan

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Answer:

6.55 mL of K₃PO₄ are required

Explanation:

We need to propose the reaction, in order to begin:

2K₃PO₄ (aq) + 3NiCl₂(aq) → Ni₃(PO₄)₂ (s) ↓ + 6KCl (aq)

Molarity = mol/L (Moles of solute that are contained in 1 L of solution.)

M = mol / volume(L). Let's find out the moles of chloride:

- We first convert the volume from mL to L → 187 mL . 1L / 1000mL = 0.187L

0.0184 M . 0.187L = 0.00344 moles of NiCl₂

Ratio is 3:2. Let's propose this rule of three:

3 moles of chloride react with 2 moles of phosphate

Then, 0.00344 moles of NiCl₂ will react with (0.00344 . 2) /3 = 0.00229 moles of K₃PO₄

M = mol / volume(L) → Volume(L) = mol/M

Volume(L) = 0.00229 mol / 0.350 M = 6.55×10⁻³L

We convert the volume from L to mL → 6.55×10⁻³L . 1000mL /1L = 6.55 mL

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3 years ago