what is the mass of the solid NHy Cl formed when 7 volu NS MIxed Nth an equal mass HCNhtis the of the gas remaingng measured at 14°c and 752 mm Hg - What ads is it? NH, (g) + HCl (g) NHy CL.CS) SH,N14.0l H:3(1.01) moles of NHz > 71.5ANH3 × -4.19 mol Acl: H : l.o1 cl:55,45 36.069 moles of ACl → 71,54 ACI = L98 moles :N 14.01 Taithing 7 HCI is Cl: 35.45 53.5
what is the mass of the solid NHy Cl formed when 7 volu NS MIxed Nth an equal mass HCNhtis the of the gas remaingng measured at 14°c and 752 mm Hg - What ads is it? NH, (g) + HCl (g) NHy CL.CS) SH,N14.0l H:3(1.01) moles of NHz > 71.5ANH3 × -4.19 mol Acl: H : l.o1 cl:55,45 36.069 moles of ACl → 71,54 ACI = L98 moles :N 14.01 Taithing 7 HCI is Cl: 35.45 53.5
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I was hoping someone could explain how to find the number of moles of NH4Cl in the attached problem? I am certain I can figure out the ideal gas formula once I get that, but I am a little unclear on what the next step is from where I am at. Thank you!
![**Title: Calculating the Mass of NH₄Cl Formed and the Volume of Remaining Gas**
**Objective:**
To determine the mass of the solid NH₄Cl formed when 7.5 g of NH₃ is mixed with an equal mass of HCl. Additionally, to calculate the volume of the gas remaining, measured at 14°C and 752 mmHg, and identify the gas.
**Chemical Reaction:**
\[ \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) \]
**Given:**
- Mass of NH₃ = 7.5 g
- Mass of HCl = 7.5 g
**1. Moles Calculation:**
- Molar mass of NH₃:
\[ N: 14.01, \; H: 3(1.01) = 17.04 \, \text{g/mol} \]
\[ \frac{7.5 \, \text{g NH}_3}{17.04 \, \text{g/mol}} = 0.44 \, \text{moles NH}_3 \]
- Molar mass of HCl:
\[ H: 1.01, \; Cl: 35.45 = 36.46 \, \text{g/mol} \]
\[ \frac{7.5 \, \text{g HCl}}{36.46 \, \text{g/mol}} = 0.21 \, \text{moles HCl} \]
**2. Limiting Reactant:**
- 1:1 stoichiometric ratio indicates HCl is the limiting reactant.
**3. Mass Calculation for NH₄Cl:**
- Molar mass of NH₄Cl:
\[ N: 14.01, \; H: 4(1.01), \; Cl: 35.45 = 53.5 \, \text{g/mol} \]
- Mass of NH₄Cl:
\[ 0.21 \, \text{moles} \times 53.5 \, \text{g/mol} = 11.24 \, \text{g NH}_4\text{Cl} \]
**Diagrams and Calculations:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8854ffd-431b-44a2-b47e-daec534c3a26%2F54d3ae6e-fc5c-4ca0-bf5c-9295589b2241%2F88lfmk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the Mass of NH₄Cl Formed and the Volume of Remaining Gas**
**Objective:**
To determine the mass of the solid NH₄Cl formed when 7.5 g of NH₃ is mixed with an equal mass of HCl. Additionally, to calculate the volume of the gas remaining, measured at 14°C and 752 mmHg, and identify the gas.
**Chemical Reaction:**
\[ \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) \]
**Given:**
- Mass of NH₃ = 7.5 g
- Mass of HCl = 7.5 g
**1. Moles Calculation:**
- Molar mass of NH₃:
\[ N: 14.01, \; H: 3(1.01) = 17.04 \, \text{g/mol} \]
\[ \frac{7.5 \, \text{g NH}_3}{17.04 \, \text{g/mol}} = 0.44 \, \text{moles NH}_3 \]
- Molar mass of HCl:
\[ H: 1.01, \; Cl: 35.45 = 36.46 \, \text{g/mol} \]
\[ \frac{7.5 \, \text{g HCl}}{36.46 \, \text{g/mol}} = 0.21 \, \text{moles HCl} \]
**2. Limiting Reactant:**
- 1:1 stoichiometric ratio indicates HCl is the limiting reactant.
**3. Mass Calculation for NH₄Cl:**
- Molar mass of NH₄Cl:
\[ N: 14.01, \; H: 4(1.01), \; Cl: 35.45 = 53.5 \, \text{g/mol} \]
- Mass of NH₄Cl:
\[ 0.21 \, \text{moles} \times 53.5 \, \text{g/mol} = 11.24 \, \text{g NH}_4\text{Cl} \]
**Diagrams and Calculations:**
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