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Elena-2011 [213]
3 years ago
10

If 2.0 ml of 6.0m hcl is used to make a 500.0-ml aqueous solution, what is the molarity of the dilute solution?

Chemistry
2 answers:
svp [43]3 years ago
4 0
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:

C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
natita [175]3 years ago
4 0

Answer:

0.024M

Explanation:

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2 years ago
If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the
Sedbober [7]

Answer:

Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

7 0
3 years ago
How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction
e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl  = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with  H₂SO₄

Reaction:

         2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the  H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

         2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

           1 mol                              2 mol  

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

            2NaCl           +      H₂SO₄     -------->       2HCl        +     Na₂SO₄

      2 mol (58.5 g/mol)                                    2 mol (36.5 g/mol)

              94 g                                                           73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

           94 g of NaCl  ≅ 73 g of HCl

           15.0 g of NaCl  ≅ X g of HCl

By Doing cross multiplication

           X g of HCl = 73 g x 15.0 g / 94 g

           X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

7 0
3 years ago
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