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3 years ago

Describe the chemical reaction based on the chemical equation below. Also, explain whether the equation is balanced.

Chemistry

2 answers:

trapecia [35]3 years ago

7 0

Answer:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(I)

We need 4 moles of ammonia gas (NH3) and 5 moles of oxygen gas( O2) to produce 4 moles of nitromonoxide gas (NO) and 6 moles of water (H2O)

Explanation:

Step 1: The unabalnced equation

NH3(g) + O2(g)→ NO(g) + H2O(I)

Step 2: Balancing the equation

NH3(g) + O2(g) → NO(g) + H2O(I)

On the left side we have 3x H (in NH3), on the right side we have 2x H (in H2O)

To balance the amount of H, on both sides, we have to multiply NH3 by 4 and H2O by 6

4NH3(g) + O2(g) → NO(g) + 6H2O(I)

We have 4x N on the left side (in 4NH3) and 1x N in NO on the right side. To balance the amount of N on both sides, we have to multiply NO on the right side by 4

4NH3(g) + O2(g) → 4NO(g) + 6H2O(I)

On the left side we have 2x O (in O2, on the right side we have 10x O (4x in 4NO and 6x in 6H2O). To balance the amount of O we have to multply O2 by 5. Now the equation is balanced.

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(I)

Step 3: Explanation

This means we need 4 moles of ammonia gas (NH3) and 5 moles of oxygen gas( O2) to produce 4 moles of nitromonoxide gas (NO) and 6 moles of water (H2O)

liubo4ka [24]3 years ago

3 0

Answer:

The balanced equation is: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(I)

4 moles of ammonia need 5 moles of oxygen to react, in order to produce 6 moles of water and 4 moles of nitrogen oxide.

Explanation:

NH₃(g) + O₂(g) → NO(g) + H₂O(I)

That is the unbalanced equation.

Ratio is 1:1, so 1 mol of ammonia and 1 mol of oxygen may react to produce 1 mol of NO and 1 mol of water

First of all we need to balance. Let's count the atoms:

In reagent side, we have 3 H and in product side there are 2H, we can add a 3, so now we have 6 H. Therefore, the reagent side must have 2 in front of the ammonia. In both sides we have 6 H

Now, that we added a 2 in the ammonia, we have 2N, so we have to add 2 to NO in the product side.

Let's count the oxygen in the product side, 5. We must add 5/2 in the reactant side. But chemical equations usually do not have rational numbers, that's why we multiply x2, all the stoichiometry.

The balanced equation is: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(I)

4 moles of ammonia need 5 moles of oxygen to react, in order to produce 6 moles of water and 4 moles of nitrogen oxide.

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3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)

GrogVix [38]

Explanation:

(a) The given reaction equation is as follows.

$Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)$ (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Reduction-half-reaction: $NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)$

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

$Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)$ (acidic)

(b) The given reaction equation is as follows.

$HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)$ (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: $HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)$

Oxidation-half reaction: $Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}$

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

$4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}$

Therefore, balancing the whole reaction equation in the basic medium as follows.

$H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)$

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A solution containing HCl and the weak acid HClO2 has a pH of 2.4. Enough KOH (aq) is added to the solution to increase the pH t

astraxan [27]

Answer:

$\boxed{\rm \text{[ClO$_{2}^{-}$] increases}}$

Explanation:

At the beginning, you have two reactions happening:

$\rm HCl + H$_{2}$O$ $\, \longrightarrow \,$ H$_{3}$O$^{+}$ + Cl$^{-}$\\\rm HClO$_{2}$ + H$_{2}$O$ $\, \rightleftharpoons \,$ H$_{3}$O$^{+}$ + ClO$_{2}^{-}$$

As you add KOH(aq), it does two things:

It increases the volume of the solution.
It reacts with the hydronium ions to form water.

A) The HCl is completely ionized. The Cl⁻ does not react, but it is diluted when the volume of the solution increases. [Cl⁻] decreases.

B) The KOH reacts with the H⁺ and removes it from the solution. [H⁺] decreases.

C) When all the H⁺ from the HCl has been neutralized, the KOH starts neutralizing the H⁺ from the HClO₂. According to Le Châtelier's Principle, more HClO₂ will dissociate to replace the decreased H⁺. [HClO₂] decreases.

D) As HClO₂ reacts, it forms ClO₂⁻. $\boxed{\rm \textbf{[ClO$_{2}^{-}$] increases}}$

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In a hydrogen molecule, the two hydrogen atoms are held together by a single bond with a bond energy of 436 kj/mol of hydrogen.

svetoff [14.1K]

Answer: 873 kJ of energy will be required to break two moles of hydrogen gas.

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Above chemical equation shows that two moles of hydrogen gas is reacting with one mole of oxygen gas to give two mole of water molecule.

Bond energy of H-H = 436kJ

According to reaction we have two moles hydrogen,so energy required to break the

H-H bond in two moles of hydrogen gas will be:

$2\times \text{bond energy of H-H bond}$

$2\times {436}$ = 872kJ

872kJ of energy will be required to break H-H bond in two moles of hydrogen gas.

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