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

NH3(g) + O2(g) > NO(g) + H2O(I)

what is the explanation?

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.

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)

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally