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vodka [1.7K]
3 years ago
9

how many moles of sodium are needed to react with sulfuric acid to produce 3.75 moles of sodium sulfate according to the followi

ng unbalanced reaction :

Chemistry
1 answer:
anastassius [24]3 years ago
5 0
Hello!
<span>
You'll need to react 7,5 moles of Sodium with sulfuric acid to produce 3.75 moles of sodium sulfate
</span>
First of all, you need to balance the reaction. The balanced reaction is shown below (ensuring that the Law of Conservation of Mass is met on both sides):

2Na + H₂SO₄ → Na₂SO₄ + H₂

Now, all that you have to do is to use molar equivalences in this reaction applying the coefficients to calculate the moles of Sodium that you'll need:

molesNa=3,75moles Na_{2} SO_4* \frac{2 moles Na}{1 mol Na_{2} SO_4} =7,5 moles Na 

Have a nice day!
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What is the difference between gases and nonmetal?
ExtremeBDS [4]

Answer:

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3 years ago
Consider the titration of a 20.0 mL sample of 0.500 M HCN (Ka =6.17x10-10) with 0.250 M KOH. a. (6pt) What is the initial pH? b.
Salsk061 [2.6K]

Answer:

a. pH = 4.75

b. pH = 9.20

c. pH = 8.42

d. pH = 13.53

Explanation:

This is a titration between a strong base, the KOH and a weak acid, HCN.

The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN

          HCN + H2O ⇄  H₃O⁺  +  CN⁻

Initial    0.5                      -             -

Eq.      0.5-x                    x             x

Ka =  x² / (0.5-x) = 6.17ₓ10⁻¹⁰

Ka is really small, so we can say that 0.5-x = 0.5. Then,

x² = 6.17ₓ10⁻¹⁰ . 0.5

x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]

pH = - log [H₃O⁺]  →  - log 1.75×10⁻⁵ = 4.75

b. First of all, we determine the moles of base, we are adding.

0.250 mol/L . 0.006 L = 0.0015 moles

In conclussion we have 0.0015 moles of OH⁻

Now, we determine the moles of our acid.

0.500 mol/L . 0.020L = 0.01 moles

The  0.0015 moles of OH⁻ will be neutralized with the acid, so:

      HCN     +    OH⁻         →     H₂O   +    CN⁻

       0.01         0.0015                          0.0085

The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)

Our new volume is 20 mL and 6mL that we added, so, 26mL

This is a buffer with the weak acid, and its conjugate base.

Our concentrations are 0.0085 moles / 0.026 L = 0.327 M

We apply Henderson-Hasselbach

pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)

pH = pKa

c. When we add 40 mL, our volume is 20mL +40mL  = 60 mL

These are the moles, we add:

0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)

 HCN     +    OH⁻         →     H₂O   +    CN⁻

  0.01          0.01                                 0.01

All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.

0.01 moles / 0.060 L = 0.16 M → [CN⁻]

pH at this point will be

       CN⁻  +  H₂O ⇄  HCN + OH⁻             Kb = 1.62ₓ10⁻⁵ (Kw/Ka)

In.   0.16                        -          -

Eq. 0.16-x                     x          x

Kb = x² / (0.16-x)

We can also assume that 0.16-x = 0.16. Then:

[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ .  0.16) = 2.59×10⁻⁶

- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58

pH = 14 - pOH  → 14 - 5.58 = 8.42

This is a basic pH, because the titration is between a weak acid and a strong base.

d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL

We add 0.5 mol/L . 0.062L = 0.031 moles

These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.

0.031 moles - 0.01 moles = 0.021 moles of OH⁻

[OH⁻] = 0.021 moles / 0.062L = 0.34M

- log [OH⁻]  = pOH → - log 0.34 = 0.47

pH = 14-pH → 14 - 0.47 = 13.53

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Good morning guys have a good day and please take care ❤️​
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Explanation:

Thank u

have good day ahead:)

4 0
3 years ago
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