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3 years ago

KOH (aq) + HNO(little3) (aq) = KNO(little 3) (aq) + H20 (1)

Chemistry

1 answer:

Yuki888 [10]3 years ago

5 0

The reaction already balanced

<h3>Further explanation </h3>

Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:

• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.

• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product

• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C, H and the last O

Reaction

KOH (aq) + HNO₃ (aq) ⇒ KNO₃ (aq) + H₂0 (l)

give coefficient

aKOH (aq) + bHNO₃ (aq) ⇒ KNO₃ (aq) + cH₂0 (l)

K : left = a, right = 1⇒a=1

N : left = b, right =1, b=1

H: left = a, right a+b=2c⇒1+1=2c⇒2=2c⇒c=1

KOH (aq) + HNO₃ (aq) ⇒ KNO₃ (aq) + H₂0 (l)⇒ already balanced

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3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

3 years ago