Answer:
**Solution:**
**4.**
**Molar masses:**
* Zn: 65.38 g/mol
* Cl: 35.45 g/mol
* Ag: 107.9 g/mol
**Reaction:**
```
ZnCl2 + 2 AgNO3 -> Zn(NO3)2 + 2 AgCl
```
**Calculations:**
1. Calculate the moles of AgCl produced:
```
moles of AgCl = 1.58 g / 143.32 g/mol = 0.011 moles
```
2. Calculate the moles of ZnCl2 in the original mixture:
```
moles of ZnCl2 = 0.011 moles AgCl * 1 mole ZnCl2 / 2 moles AgCl = 0.0055 moles
```
3. Calculate the mass of ZnCl2 in the original mixture:
```
mass of ZnCl2 = 0.0055 moles * 136.29 g/mol = 0.750 g
```
4. Calculate the percentage by mass of ZnCl2 in the original mixture:
```
percentage by mass of ZnCl2 = (0.750 g / 1.33 g) * 100% = 56.3%
```
**Therefore, the percentage by mass of ZnCl2 in the original mixture is 56.3%.**
**5.**
**Molar masses:**
* C: 12.01 g/mol
* H: 1.008 g/mol
* O: 16.00 g/mol
**Reaction:**
```
CH4 + 2 O2 -> CO2 + 2 H2O
```
**Calculations:**
1. Calculate the moles of CO2 produced:
```
moles of CO2 = 2.039 g / 44.01 g/mol = 0.0461 moles
```
2. Calculate the moles of CH4 in the original mixture:
```
moles of CH4 = 0.0461 moles CO2 * 1 mole CH4 / 1 mole CO2 = 0.0461 moles
```
3. Calculate the mass of CH4 in the original mixture:
```
mass of CH4 = 0.0461 moles * 16.04 g/mol = 0.740 g
```
4. Calculate the percentage by mass of CH4 in the original mixture:
```
percentage by mass of CH4 = (0.740 g / 0.732 g) * 100% = 101%
```
**Therefore, the percentage by mass of CH4 in the original mixture is 101%.**
**(Note: The answer is greater than 100% because the question states that the mixture is burned in excess oxygen. This means that all of the carbon in the methane and ethane is converted into carbon dioxide, even if there is more carbon than oxygen in the mixture.)**
**6.**
**Molar masses:**
* HCl: 36.458 g/mol
* H2O: 18.016 g/mol
**Calculations:**
1. Calculate the mass of HCl in the stock solution:
```
mass of HCl = 36.0% * 118 g/mL = 42.5 g
```
2. Calculate the volume of stock solution required:
```
volume of stock solution = 42.5 g / 1.18 g/mL = 35.9 mL
```
**Therefore, the volume of stock solution required is 35.9 mL.**
**7.**
**Molar masses:**
* H2SO4: 98.079 g/mol
* KOH: 56.106 g/mol
**Reaction:**
```
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
```
**Calculations:**
1. Calculate the moles of KOH in the solution:
```
moles of KOH = 0.0493 L * 0.830 mol/L = 0.0410 moles
```
2. Calculate the moles of H2SO4 required to neutralize the KOH:
```
moles of H2SO4 = 0.0410 moles KOH * 1 mole H2SO4 / 2 moles KOH = 0.0205 moles
```
3. Calculate the volume of 1.33 mol/L H2SO4 required to neutralize the KOH:
```
volume of H2SO4 = 0.0205 moles * 1 L / 1.33 mol = 0.0154 L
```
4. Convert the volume to milliliters:
```
volume of H2SO4 = 0.0154 L * 1000 mL/L = 15.4 mL
```
**Therefore, the volume of 1.33 mol L−1 H2SO4(aq) required to completely neutralize 49.3 mL of 0.830 mol L−1 KOH(aq) is 15.4 mL.**
**8.**
**Molar masses:**
* ZnCl2: 136.29 g/mol
* Zn: 65.38 g/mol
* HCl: 36.46 g/mol
**Reaction:**
```
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
```
**Calculations:**
1. Calculate the mass of atoms in the desired product:
mass of atoms in ZnCl2 = (136.29 g/mol ZnCl2) * (1 atom Zn/molecule + 2 atoms Cl/molecule) = 272.58 g/mol
2. Calculate the mass of atoms in all of the reactants:
mass of atoms in reactants = (65.38 g/mol Zn) * 1 atom Zn/molecule + 2 * (36.46 g/mol HCl) * (1 atom H/molecule + 1 atom Cl/molecule)
= 188.28 g/mol
3. Calculate the percent atom economy:
```
% AE = (272.58 g/mol ZnCl2 / 188.28 g/mol reactants) * 100% = 98.54%
```
**Therefore, the percent atom economy (% AE) for forming ZnCl2(aq) is 98.54%.**
I hope this helps!.