How many grams of phosphoric acid (97.99 g/mol) can be produced from the reaction of 10.0 g o P4010 (283.9 g/mol) with 5.00 grams of water (18.02 g/mol) according to the following balanced reaction? P4010(s) + 6 H₂O(l) → 4 H3PO4(aq) 15.0 g 18.1 g 13.8 g

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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### Chemical Reaction Problem

**Question:**
How many grams of phosphoric acid (97.99 g/mol) can be produced from the reaction of 10.0 g of P\(_4\)O\(_{10}\) (283.9 g/mol) with 5.00 grams of water (18.02 g/mol) according to the following balanced reaction?

\[ \text{P}_4\text{O}_{10(s)} + 6 \text{H}_2\text{O}_{(l)} \rightarrow 4 \text{H}_3\text{PO}_{4(aq)} \]

**Options:**
- 15.0 g
- 18.1 g
- 13.8 g

### Explanation:

To solve this, follow these steps:

1. **Calculate the moles of P\(_4\)O\(_{10}\) and H\(_2\)O**:
   - Moles of P\(_4\)O\(_{10}\) = \(\frac{10.0 \ \text{g}}{283.9 \ \text{g/mol}}\)
   - Moles of H\(_2\)O = \(\frac{5.00 \ \text{g}}{18.02 \ \text{g/mol}}\)

2. **Determine the limiting reagent**:
   - The balanced chemical equation shows that 1 mole of P\(_4\)O\(_{10}\) reacts with 6 moles of H\(_2\)O to produce 4 moles of H\(_3\)PO\(_4\).

3. **Use the stoichiometry of the reaction to find the moles of H\(_3\)PO\(_4\)**:
   - Using the moles of the limiting reagent, calculate the moles of H\(_3\)PO\(_4\) produced.

4. **Convert moles of H\(_3\)PO\(_4\) to grams**:
   - Grams of H\(_3\)PO\(_4\) = moles of H\(_3\)PO\(_4\) \(\times\) molar mass of H\(_3\)PO\(_4\) (97.99 g/mol).

By working through these steps, you can determine
Transcribed Image Text:### Chemical Reaction Problem **Question:** How many grams of phosphoric acid (97.99 g/mol) can be produced from the reaction of 10.0 g of P\(_4\)O\(_{10}\) (283.9 g/mol) with 5.00 grams of water (18.02 g/mol) according to the following balanced reaction? \[ \text{P}_4\text{O}_{10(s)} + 6 \text{H}_2\text{O}_{(l)} \rightarrow 4 \text{H}_3\text{PO}_{4(aq)} \] **Options:** - 15.0 g - 18.1 g - 13.8 g ### Explanation: To solve this, follow these steps: 1. **Calculate the moles of P\(_4\)O\(_{10}\) and H\(_2\)O**: - Moles of P\(_4\)O\(_{10}\) = \(\frac{10.0 \ \text{g}}{283.9 \ \text{g/mol}}\) - Moles of H\(_2\)O = \(\frac{5.00 \ \text{g}}{18.02 \ \text{g/mol}}\) 2. **Determine the limiting reagent**: - The balanced chemical equation shows that 1 mole of P\(_4\)O\(_{10}\) reacts with 6 moles of H\(_2\)O to produce 4 moles of H\(_3\)PO\(_4\). 3. **Use the stoichiometry of the reaction to find the moles of H\(_3\)PO\(_4\)**: - Using the moles of the limiting reagent, calculate the moles of H\(_3\)PO\(_4\) produced. 4. **Convert moles of H\(_3\)PO\(_4\) to grams**: - Grams of H\(_3\)PO\(_4\) = moles of H\(_3\)PO\(_4\) \(\times\) molar mass of H\(_3\)PO\(_4\) (97.99 g/mol). By working through these steps, you can determine
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