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3 years ago

Identify the missing coefficient in the balanced equation and classify the type of reaction.

Chemistry

2 answers:

Lostsunrise [7]3 years ago

5 0

Answer:

Option 6 ) Neutralization

Explanation:

For this case, the missing coefficient would be a "6" before the H₂O, within final products (right side of the equiation), hence, the final reaction should be:

2H₃PO₄ + 3Ba(OH)₂ ------> Ba₃(PO₄)₂ + 6H₂O

You should have in mind that the amount of atoms at each side of the chemical equation should be the same, so as to comply with the principle of mass conservation. If you add "6" on the left side of the H₂O, the equation will be balanced (for each side, lef and right, you will have: 12H, 2P, 14O and 3Ba)

Lastly, this is a chemical neutralization reaction, where an acid (H₃PO₄) is reacting with a base (Ba(OH)₂) in order to finally obtain a neutral salt (Ba₃(PO₄)₂) and water (H₂O)

solong [7]3 years ago

4 0

Answer:

6; Neutralization

Explanation:

$2H_3 PO_4+3Ba(OH)_2>Ba_3 (PO_4)_2+H_2 O$

(Unbalanced)

Balancing is making the number of atoms of each element same on both the sides (reactant and product side).

To find the number of atoms of each element we multiply coefficient and the subscript

For example $5Ca_1Cl_2$ contains

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example $3Ca_3 (P_1 O_4 )_2$ we multiply coefficient × subscript × number outside the bracket to find the number of atoms

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

3 × 3 = 9 , Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present

Reactant side Number of atoms or groups Product side

of each element

3 Ba 3

2 $PO_4$ 2

12 H 2

6 O (Exclude O of $PO_4$ ) 1

To balance H and O, multiply $H_2O$ by 6 so we have

$2H_3 PO_4+3Ba(OH)_2>Ba_3 (PO_4)_2+6 H_2 O$

Reactant side Number of atoms or groups Product side

of each element

3 Ba 3

2 $PO_4$ 2

12 H 12

6 O (Exclude O of $PO_4$ ) 6

Balanced!!!!

Acid and base reacts to form salt and water is called a neutralization reaction.

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Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be

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Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

$\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles$

$\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

$\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles$

The concentration of benzoic acid is:

$C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M$

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺

0.14 - x x x

$Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}$

$Ka = \frac{x*x}{0.14 - x}$

$6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0$ (2)

By solving equation (2) for x we have:

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

$pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52$

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻ (3)

The number of moles of C₆H₅CO₂⁻ is:

$\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles$