How many liters of 0.305 M K.PO. solution are necessary to completely react with 187 mL of 0.0184 M NICI according to the balanced chemical reaction: KPO:(aq) + 3 NICI:(aq) Nis(PO.):(s) + KCI(aq) molNICIE 187 mL NICI2 2 LKPO. 0.0184 L NICI2 3 mo-NICIE 0.305 mol KPO. 0.0184 mol NiCl2 187 mL NICI2 2L KPO4 L NICI: 3 mol NiCl2 0.305 mol KPO. ADD FACTOR DELETE ANSWER RESET *( ) 0.0113 6.022 x 1023 0.305 187 2 1000 0.001 3 6 0.00376 0.00752 0.0184 7.52 mL K.PO. M К.РО M NICI2 g KPO. LKPO. mol KPO. mol NiCl2 mL NICI2 L NICI g NiCl2
How many liters of 0.305 M K.PO. solution are necessary to completely react with 187 mL of 0.0184 M NICI according to the balanced chemical reaction: KPO:(aq) + 3 NICI:(aq) Nis(PO.):(s) + KCI(aq) molNICIE 187 mL NICI2 2 LKPO. 0.0184 L NICI2 3 mo-NICIE 0.305 mol KPO. 0.0184 mol NiCl2 187 mL NICI2 2L KPO4 L NICI: 3 mol NiCl2 0.305 mol KPO. ADD FACTOR DELETE ANSWER RESET *( ) 0.0113 6.022 x 1023 0.305 187 2 1000 0.001 3 6 0.00376 0.00752 0.0184 7.52 mL K.PO. M К.РО M NICI2 g KPO. LKPO. mol KPO. mol NiCl2 mL NICI2 L NICI g NiCl2
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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I need help with this equation. Please show me how to organize the equation, just EXACT like the screenshot. I knew the final result is 0.00752 L
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![### Stoichiometry Problem: Calculating Solution Volume
**Problem Statement:**
How many liters of a 0.305 M K₃PO₄ solution are necessary to completely react with 187 mL of a 0.0184 M NiCl₂ solution, according to the balanced chemical reaction:
\[
2 \text{ K}_3\text{PO}_4(\text{aq}) + 3 \text{ NiCl}_2(\text{aq}) \rightarrow \text{ Ni}_3(\text{PO}_4)_2(\text{s}) + 6 \text{ KCl}(\text{aq})
\]
**Solution Steps:**
1. **Initial Information:**
- Molarity of NiCl₂: 0.0184 mol/L
- Volume of NiCl₂: 187 mL
- Molarity of K₃PO₄: 0.305 mol/L
2. **Conversion Process:**
- Convert mL to L:
\[
187 \text{ mL NiCl}_2 \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.187 \text{ L}
\]
- Determine moles of NiCl₂:
\[
0.187 \text{ L NiCl}_2 \times 0.0184 \text{ mol/L} = 0.0034368 \text{ mol NiCl}_2
\]
- Use stoichiometry to find moles of K₃PO₄:
\[
0.0034368 \text{ mol NiCl}_2 \times \frac{2 \text{ mol K}_3\text{PO}_4}{3 \text{ mol NiCl}_2} = 0.0022912 \text{ mol K}_3\text{PO}_4
\]
- Calculate volume of K₃PO₄ required:
\[
0.0022912 \text{ mol K}_3\text{PO}_4 \times \frac{1 \text{ L}}{0.305 \text{ mol}} \approx 0.00752 \text{ L}
\]
\( \approx 7.52 \text{ mL} \)
**Interactive Interface:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F048127f3-de9a-44a1-9086-2f981c8bd58d%2Ff6502596-5ffe-48c0-bb45-7be59d5fe068%2Fd4xc73_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Stoichiometry Problem: Calculating Solution Volume
**Problem Statement:**
How many liters of a 0.305 M K₃PO₄ solution are necessary to completely react with 187 mL of a 0.0184 M NiCl₂ solution, according to the balanced chemical reaction:
\[
2 \text{ K}_3\text{PO}_4(\text{aq}) + 3 \text{ NiCl}_2(\text{aq}) \rightarrow \text{ Ni}_3(\text{PO}_4)_2(\text{s}) + 6 \text{ KCl}(\text{aq})
\]
**Solution Steps:**
1. **Initial Information:**
- Molarity of NiCl₂: 0.0184 mol/L
- Volume of NiCl₂: 187 mL
- Molarity of K₃PO₄: 0.305 mol/L
2. **Conversion Process:**
- Convert mL to L:
\[
187 \text{ mL NiCl}_2 \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.187 \text{ L}
\]
- Determine moles of NiCl₂:
\[
0.187 \text{ L NiCl}_2 \times 0.0184 \text{ mol/L} = 0.0034368 \text{ mol NiCl}_2
\]
- Use stoichiometry to find moles of K₃PO₄:
\[
0.0034368 \text{ mol NiCl}_2 \times \frac{2 \text{ mol K}_3\text{PO}_4}{3 \text{ mol NiCl}_2} = 0.0022912 \text{ mol K}_3\text{PO}_4
\]
- Calculate volume of K₃PO₄ required:
\[
0.0022912 \text{ mol K}_3\text{PO}_4 \times \frac{1 \text{ L}}{0.305 \text{ mol}} \approx 0.00752 \text{ L}
\]
\( \approx 7.52 \text{ mL} \)
**Interactive Interface:
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