Consider the following equation: mol of ammopiut. be (tuboy) 3 AgNO3(aq) + Na3PO4(aq) Ag3PO4(s) + 3 NaNO3(aq) When 15.3 g of AgNO3(aq) is mixed with 10.9 g of Na3PO4(aq), how many grams of Ag3PO4(s) can be produced? a. 10.9 g b. 37.7 g c. 40.4 g d. 27.8 g e. 12.6 g
Consider the following equation: mol of ammopiut. be (tuboy) 3 AgNO3(aq) + Na3PO4(aq) Ag3PO4(s) + 3 NaNO3(aq) When 15.3 g of AgNO3(aq) is mixed with 10.9 g of Na3PO4(aq), how many grams of Ag3PO4(s) can be produced? a. 10.9 g b. 37.7 g c. 40.4 g d. 27.8 g e. 12.6 g
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![### Question
17. Consider the following equation:
\[ 3 \text{AgNO}_3(\text{aq}) + \text{Na}_3\text{PO}_4(\text{aq}) \longrightarrow \text{Ag}_3\text{PO}_4(\text{s}) + 3 \text{NaNO}_3(\text{aq}) \]
When 15.3 g of AgNO₃(aq) is mixed with 10.9 g of Na₃PO₄(aq), how many grams of Ag₃PO₄(s) can be produced?
### Options
a. 10.9 g
b. 37.7 g
c. 40.4 g
d. 27.8 g
e. 12.6 g
### Explanation
This problem involves a chemical reaction and requires the calculation of the mass of a product formed from given amounts of reactants. Here's a step-by-step breakdown to find the solution:
1. **Determine the molar masses of the reactants and the product:**
- **Molecular weight of AgNO₃:**
- Ag = 107.87 g/mol
- N = 14.01 g/mol
- O = 16.00 g/mol (3 atoms of O)
- Total = 107.87 + 14.01 + (3 × 16.00) = 169.88 g/mol
- **Molecular weight of Na₃PO₄:**
- Na = 22.99 g/mol (3 atoms of Na)
- P = 30.97 g/mol
- O = 16.00 g/mol (4 atoms of O)
- Total = (3 × 22.99) + 30.97 + (4 × 16.00) = 163.94 g/mol
- **Molecular weight of Ag₃PO₄:**
- Ag = 107.87 g/mol (3 atoms of Ag)
- P = 30.97 g/mol
- O = 16.00 g/mol (4 atoms of O)
- Total = (3 × 107.87) + 30.97 + (4 × 16.00) = 418.58 g/mol
2. **Calculate the moles of each react](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F066bb5df-9069-4300-b122-0f92c568ce58%2Fc4da2978-1027-4965-bce2-ce3bbe9c02de%2Fzii6alf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question
17. Consider the following equation:
\[ 3 \text{AgNO}_3(\text{aq}) + \text{Na}_3\text{PO}_4(\text{aq}) \longrightarrow \text{Ag}_3\text{PO}_4(\text{s}) + 3 \text{NaNO}_3(\text{aq}) \]
When 15.3 g of AgNO₃(aq) is mixed with 10.9 g of Na₃PO₄(aq), how many grams of Ag₃PO₄(s) can be produced?
### Options
a. 10.9 g
b. 37.7 g
c. 40.4 g
d. 27.8 g
e. 12.6 g
### Explanation
This problem involves a chemical reaction and requires the calculation of the mass of a product formed from given amounts of reactants. Here's a step-by-step breakdown to find the solution:
1. **Determine the molar masses of the reactants and the product:**
- **Molecular weight of AgNO₃:**
- Ag = 107.87 g/mol
- N = 14.01 g/mol
- O = 16.00 g/mol (3 atoms of O)
- Total = 107.87 + 14.01 + (3 × 16.00) = 169.88 g/mol
- **Molecular weight of Na₃PO₄:**
- Na = 22.99 g/mol (3 atoms of Na)
- P = 30.97 g/mol
- O = 16.00 g/mol (4 atoms of O)
- Total = (3 × 22.99) + 30.97 + (4 × 16.00) = 163.94 g/mol
- **Molecular weight of Ag₃PO₄:**
- Ag = 107.87 g/mol (3 atoms of Ag)
- P = 30.97 g/mol
- O = 16.00 g/mol (4 atoms of O)
- Total = (3 × 107.87) + 30.97 + (4 × 16.00) = 418.58 g/mol
2. **Calculate the moles of each react
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