I’ve solved the top portion, I need help solving the bottom portion. Would you break it down for me? Thank you. # Drill on Reactants & Products**Objective:** Fill in all of the rectangles in each row that correspond to the one quantity given in that row for the reaction: \[ \text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \text{H}_2\text{O} \] You may be asked to turn in this sheet.| Reacting | Forming ||--------------------------------------------|--------------------------------------------|| moles NH₃ | g NH₃ | moles O₂ | g O₂ | moles NO | g NO | moles H₂O | g H₂O | molecules H₂O || --------- | ----- | -------- | ---- | -------- | ---- | --------- | ----- | -------------- || 1.00 | | | | | | | | || | 85.0 | | | | | | | || | | 0.600 | | | | | | || | | | | 0.320 | | | | || | | | | | 11.5 | | | || | | | | | | | | 9.0 × 10²⁴ | Reacting Forming moles NH3 g NH3 moles O2 g O2 moles NO g NO moles H2O g H2O molecules H2O 1.00 17.0 1.25 40.0 1.00 30.0 1.50 27.0 9.03 x 1023 5.00 85.0 6.25 200.0 5.00 150.0 7.50 135 4.52 x 1024 0.480 8.16 0.600 19.2 0.480 14.4 0.720 <

Answered: Fill in all of the rectangles in each…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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I’ve solved the top portion, I need help solving the bottom portion. Would you break it down for me? Thank you.

# Drill on Reactants & Products

**Objective:** Fill in all of the rectangles in each row that correspond to the one quantity given in that row for the reaction:
\[ \text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \text{H}_2\text{O} \]
You may be asked to turn in this sheet.

| Reacting | Forming |
|--------------------------------------------|--------------------------------------------|
| moles NH₃ | g NH₃ | moles O₂ | g O₂ | moles NO | g NO | moles H₂O | g H₂O | molecules H₂O |
| --------- | ----- | -------- | ---- | -------- | ---- | --------- | ----- | -------------- |
| 1.00 | | | | | | | | |
| | 85.0 | | | | | | | |
| | | 0.600 | | | | | | |
| | | | | 0.320 | | | | |
| | | | | | 11.5 | | | |
| | | | | | | | | 9.0 × 10²⁴ |

<table>
<thead>
<tr>
<th colspan="5">Reacting</th>
<th colspan="5">Forming</th>
</tr>
<tr>
<th>moles NH<sub>3</sub></th>
<th>g NH<sub>3</sub></th>
<th>moles O<sub>2</sub></th>
<th>g O<sub>2</sub></th>
<th></th>
<th>moles NO</th>
<th>g NO</th>
<th>moles H<sub>2</sub>O</th>
<th>g H<sub>2</sub>O</th>
<th>molecules H<sub>2</sub>O</th>
</tr>
</thead>
<tbody>
<tr>
<td>1.00</td>
<td>17.0</td>
<td>1.25</td>
<td>40.0</td>
<td></td>
<td>1.00</td>
<td>30.0</td>
<td>1.50</td>
<td>27.0</td>
<td>9.03 x 10<sup>23</sup></td>
</tr>
<tr>
<td>5.00</td>
<td>85.0</td>
<td>6.25</td>
<td>200.0</td>
<td></td>
<td>5.00</td>
<td>150.0</td>
<td>7.50</td>
<td>135</td>
<td>4.52 x 10<sup>24</sup></td>
</tr>
<tr>
<td>0.480</td>
<td>8.16</td>
<td>0.600</td>
<td>19.2</td>
<td></td>
<td>0.480</td>
<td>14.4</td>
<td>0.720</td>
<

Expert Solution

Step 1: Balanced chemical equation

To identify the stoichiometric amounts of the reactants and products present in a given system, the balanced chemical equation, in which both sides of the equation have same number of atoms, must be determined.

The skeletal equation for the given reaction is-

$NH subscript 3 plus straight O subscript 2 rightwards arrow NO plus straight H subscript 2 straight O$

In this, the number of all atoms except H, is balanced on both sides of the equation. To balance the H atom, 3/2 is added as a coefficient before H₂O on the product side as follows-

$NH subscript 3 plus straight O subscript 2 rightwards arrow NO plus 3 over 2 straight H subscript 2 straight O$

Next, to balance the number of O atoms in the above equation, add 5/4 as the coefficient before O₂ on the reactant side as follows-

$NH subscript 3 plus 5 over 4 straight O subscript 2 rightwards arrow NO plus 3 over 2 straight H subscript 2 straight O$

Finally, the above equation is multiplied by 4 to obtain whole number coefficients as follows-

$4 NH subscript 3 plus 5 straight O subscript 2 rightwards arrow 4 NO plus 6 straight H subscript 2 straight O$

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