If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl th recipitates? Express your answer in decimal notation rounded to three significant figures. CaCl₂(aq) + 2AgNO3 (aq) → 2AgCl(s) + Ca(NO3)2(aq) MM of AgCl = 143.5 g/mol
If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl th recipitates? Express your answer in decimal notation rounded to three significant figures. CaCl₂(aq) + 2AgNO3 (aq) → 2AgCl(s) + Ca(NO3)2(aq) MM of AgCl = 143.5 g/mol
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![### Precipitation Reaction: Calculating the Mass of AgCl Formed
#### Problem Statement:
If 30.0 mL of 0.150 M CaCl₂ is added to 15.0 mL of 0.100 M AgNO₃, what is the mass in grams of AgCl that precipitates? Express your answer in decimal notation rounded to three significant figures.
#### Reaction Equation:
\[ \text{CaCl}_2(\text{aq}) + 2\text{AgNO}_3 (\text{aq}) \rightarrow 2\text{AgCl}(\text{s}) + \text{Ca(NO}_3\text{)}_2(\text{aq}) \]
#### Molar Mass:
Molecular Mass (MM) of AgCl = 143.5 g/mol
### Steps:
1. **Calculate the moles of each reactant:**
- Moles of CaCl₂: \( \frac{0.0300 \text{ L} \times 0.150 \text{ M}} = 0.00450 \text{ moles} \)
- Moles of AgNO₃: \( \frac{0.0150 \text{ L} \times 0.100 \text{ M}} = 0.00150 \text{ moles} \)
2. **Determine the limiting reactant:**
- From the balanced equation: 1 mole of CaCl₂ reacts with 2 moles of AgNO₃.
- Required moles of AgNO₃ for 0.00450 moles of CaCl₂:
\( 0.00450 \text{ moles} \times 2 = 0.00900 \text{ moles} \)
- Since we only have 0.00150 moles of AgNO₃, AgNO₃ is the limiting reactant.
3. **Calculate the moles of AgCl produced:**
- From the balanced equation, 2 moles of AgNO₃ produce 2 moles of AgCl.
- Therefore, 0.00150 moles of AgNO₃ will produce 0.00150 moles of AgCl.
4. **Calculate the mass of AgCl precipitated:**
- Mass of AgCl = moles of AgCl × molar mass of AgCl](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9413ee1b-018f-4e4f-9457-fb88f0adac84%2F02932316-17bc-40d7-90a0-126e2e28d85c%2Fbk2gaw8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Precipitation Reaction: Calculating the Mass of AgCl Formed
#### Problem Statement:
If 30.0 mL of 0.150 M CaCl₂ is added to 15.0 mL of 0.100 M AgNO₃, what is the mass in grams of AgCl that precipitates? Express your answer in decimal notation rounded to three significant figures.
#### Reaction Equation:
\[ \text{CaCl}_2(\text{aq}) + 2\text{AgNO}_3 (\text{aq}) \rightarrow 2\text{AgCl}(\text{s}) + \text{Ca(NO}_3\text{)}_2(\text{aq}) \]
#### Molar Mass:
Molecular Mass (MM) of AgCl = 143.5 g/mol
### Steps:
1. **Calculate the moles of each reactant:**
- Moles of CaCl₂: \( \frac{0.0300 \text{ L} \times 0.150 \text{ M}} = 0.00450 \text{ moles} \)
- Moles of AgNO₃: \( \frac{0.0150 \text{ L} \times 0.100 \text{ M}} = 0.00150 \text{ moles} \)
2. **Determine the limiting reactant:**
- From the balanced equation: 1 mole of CaCl₂ reacts with 2 moles of AgNO₃.
- Required moles of AgNO₃ for 0.00450 moles of CaCl₂:
\( 0.00450 \text{ moles} \times 2 = 0.00900 \text{ moles} \)
- Since we only have 0.00150 moles of AgNO₃, AgNO₃ is the limiting reactant.
3. **Calculate the moles of AgCl produced:**
- From the balanced equation, 2 moles of AgNO₃ produce 2 moles of AgCl.
- Therefore, 0.00150 moles of AgNO₃ will produce 0.00150 moles of AgCl.
4. **Calculate the mass of AgCl precipitated:**
- Mass of AgCl = moles of AgCl × molar mass of AgCl
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