Answer :
Let's tackle the given problems one by one. Bear in mind that the answer from the subsequent steps regarding the gold question has been determined previously.
### Problem 1: Zinc Displacing Silver from Silver Nitrate
The balanced chemical equation is:
[tex]\[ \text{Zn} (s) + 2 \text{AgNO}_3 (aq) \rightarrow \text{Zn(NO}_3\text{)}_2 (aq) + 2 \text{Ag} (s) \][/tex]
We need to find the mass of zinc required to displace 60 g of silver.
1. Calculate Moles of Silver:
The molar mass of silver (Ag) is approximately 107.87 g/mol.
[tex]\[ \text{Moles of Ag} = \frac{60 \, \text{g}}{107.87 \, \text{g/mol}} \approx 0.556 \, \text{mol} \][/tex]
2. Determine Moles of Zinc Needed:
According to the balanced equation, 1 mole of Zn displaces 2 moles of Ag. Therefore:
[tex]\[ \text{Moles of Zn} = \frac{0.556 \, \text{mol Ag}}{2} \approx 0.278 \, \text{mol Zn} \][/tex]
3. Convert Moles of Zinc to Mass:
The molar mass of zinc (Zn) is approximately 65.38 g/mol.
[tex]\[ \text{Mass of Zn} = 0.278 \, \text{mol} \times 65.38 \, \text{g/mol} \approx 18.17 \, \text{g} \][/tex]
Therefore, approximately 18.17 g of zinc should be used.
### Problem 2: Magnesium Reacting with Oxygen to Form Magnesium Oxide
The balanced chemical equation is:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
Given 380 g of magnesium, we need to find the mass of magnesium oxide formed.
1. Calculate Moles of Magnesium:
The molar mass of magnesium (Mg) is approximately 24.305 g/mol.
[tex]\[ \text{Moles of Mg} = \frac{380 \, \text{g}}{24.305 \, \text{g/mol}} \approx 15.63 \, \text{mol} \][/tex]
2. Determine Moles of Magnesium Oxide Produced:
According to the balanced equation, 2 moles of Mg produce 2 moles of MgO. Therefore:
[tex]\[ \text{Moles of MgO} = 15.63 \, \text{mol} \][/tex]
3. Convert Moles of Magnesium Oxide to Mass:
The molar mass of magnesium oxide (MgO) is approximately 40.304 g/mol.
[tex]\[ \text{Mass of MgO} = 15.63 \, \text{mol} \times 40.304 \, \text{g/mol} \approx 629.75 \, \text{g} \][/tex]
Therefore, approximately 629.75 g of magnesium oxide is formed.
### Problem 3: Aluminium Dissolved in Hydrochloric Acid Producing Hydrogen
The balanced chemical equation is:
[tex]\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \][/tex]
Given 33 g of hydrogen produced, we need to find the mass of aluminium dissolved.
1. Calculate Moles of Hydrogen:
The molar mass of hydrogen (H₂) is approximately 2.02 g/mol.
[tex]\[ \text{Moles of } \text{H}_2 = \frac{33 \, \text{g}}{2.02 \, \text{g/mol}} \approx 16.34 \, \text{mol} \][/tex]
2. Determine Moles of Aluminium Used:
According to the balanced equation, 3 moles of H₂ are produced from 2 moles of Al. Therefore:
[tex]\[ \text{Moles of Al} = \frac{16.34 \, \text{mol H}_2}{3} \times 2 \approx 10.89 \, \text{mol Al} \][/tex]
3. Convert Moles of Aluminium to Mass:
The molar mass of aluminium (Al) is approximately 26.98 g/mol.
[tex]\[ \text{Mass of Al} = 10.89 \, \text{mol} \times 26.98 \, \text{g/mol} \approx 293.95 \, \text{g} \][/tex]
Therefore, approximately 293.95 g of aluminium was dissolved.
### Problem 4: Mass of Two Moles of Potassium Dichromate (K₂Cr₂O₇)
The molar mass of potassium dichromate (K₂Cr₂O₇) can be calculated by summing the atomic masses of its constituents:
[tex]\[ \text{Molar mass of } K_2Cr_2O_7 = 2(39.10) + 2(51.9961) + 7(15.999) \approx 294.18 \, \text{g/mol} \][/tex]
For two moles:
[tex]\[ \text{Mass of 2 moles of } K_2Cr_2O_7 = 2 \times 294.18 \approx 588.36 \, \text{g} \][/tex]
Therefore, the mass of two moles of potassium dichromate is approximately 588.36 g.
### Problem 5: Number of Moles of Gold (Au)
Given 9.85 × 10⁻³ g of gold, we need to calculate the number of moles.
The molar mass of gold (Au) is approximately 196.97 g/mol.
[tex]\[ \text{Number of moles} = \frac{9.85 \times 10^{-3} \, \text{g}}{196.97 \, \text{g/mol}} \approx 5.00085 \times 10^{-5} \, \text{mol} \][/tex]
Therefore, the number of moles represented by 9.85 × 10⁻³ g of gold is approximately [tex]\(5.00085 \times 10^{-5}\)[/tex] mol.
### Problem 1: Zinc Displacing Silver from Silver Nitrate
The balanced chemical equation is:
[tex]\[ \text{Zn} (s) + 2 \text{AgNO}_3 (aq) \rightarrow \text{Zn(NO}_3\text{)}_2 (aq) + 2 \text{Ag} (s) \][/tex]
We need to find the mass of zinc required to displace 60 g of silver.
1. Calculate Moles of Silver:
The molar mass of silver (Ag) is approximately 107.87 g/mol.
[tex]\[ \text{Moles of Ag} = \frac{60 \, \text{g}}{107.87 \, \text{g/mol}} \approx 0.556 \, \text{mol} \][/tex]
2. Determine Moles of Zinc Needed:
According to the balanced equation, 1 mole of Zn displaces 2 moles of Ag. Therefore:
[tex]\[ \text{Moles of Zn} = \frac{0.556 \, \text{mol Ag}}{2} \approx 0.278 \, \text{mol Zn} \][/tex]
3. Convert Moles of Zinc to Mass:
The molar mass of zinc (Zn) is approximately 65.38 g/mol.
[tex]\[ \text{Mass of Zn} = 0.278 \, \text{mol} \times 65.38 \, \text{g/mol} \approx 18.17 \, \text{g} \][/tex]
Therefore, approximately 18.17 g of zinc should be used.
### Problem 2: Magnesium Reacting with Oxygen to Form Magnesium Oxide
The balanced chemical equation is:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
Given 380 g of magnesium, we need to find the mass of magnesium oxide formed.
1. Calculate Moles of Magnesium:
The molar mass of magnesium (Mg) is approximately 24.305 g/mol.
[tex]\[ \text{Moles of Mg} = \frac{380 \, \text{g}}{24.305 \, \text{g/mol}} \approx 15.63 \, \text{mol} \][/tex]
2. Determine Moles of Magnesium Oxide Produced:
According to the balanced equation, 2 moles of Mg produce 2 moles of MgO. Therefore:
[tex]\[ \text{Moles of MgO} = 15.63 \, \text{mol} \][/tex]
3. Convert Moles of Magnesium Oxide to Mass:
The molar mass of magnesium oxide (MgO) is approximately 40.304 g/mol.
[tex]\[ \text{Mass of MgO} = 15.63 \, \text{mol} \times 40.304 \, \text{g/mol} \approx 629.75 \, \text{g} \][/tex]
Therefore, approximately 629.75 g of magnesium oxide is formed.
### Problem 3: Aluminium Dissolved in Hydrochloric Acid Producing Hydrogen
The balanced chemical equation is:
[tex]\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \][/tex]
Given 33 g of hydrogen produced, we need to find the mass of aluminium dissolved.
1. Calculate Moles of Hydrogen:
The molar mass of hydrogen (H₂) is approximately 2.02 g/mol.
[tex]\[ \text{Moles of } \text{H}_2 = \frac{33 \, \text{g}}{2.02 \, \text{g/mol}} \approx 16.34 \, \text{mol} \][/tex]
2. Determine Moles of Aluminium Used:
According to the balanced equation, 3 moles of H₂ are produced from 2 moles of Al. Therefore:
[tex]\[ \text{Moles of Al} = \frac{16.34 \, \text{mol H}_2}{3} \times 2 \approx 10.89 \, \text{mol Al} \][/tex]
3. Convert Moles of Aluminium to Mass:
The molar mass of aluminium (Al) is approximately 26.98 g/mol.
[tex]\[ \text{Mass of Al} = 10.89 \, \text{mol} \times 26.98 \, \text{g/mol} \approx 293.95 \, \text{g} \][/tex]
Therefore, approximately 293.95 g of aluminium was dissolved.
### Problem 4: Mass of Two Moles of Potassium Dichromate (K₂Cr₂O₇)
The molar mass of potassium dichromate (K₂Cr₂O₇) can be calculated by summing the atomic masses of its constituents:
[tex]\[ \text{Molar mass of } K_2Cr_2O_7 = 2(39.10) + 2(51.9961) + 7(15.999) \approx 294.18 \, \text{g/mol} \][/tex]
For two moles:
[tex]\[ \text{Mass of 2 moles of } K_2Cr_2O_7 = 2 \times 294.18 \approx 588.36 \, \text{g} \][/tex]
Therefore, the mass of two moles of potassium dichromate is approximately 588.36 g.
### Problem 5: Number of Moles of Gold (Au)
Given 9.85 × 10⁻³ g of gold, we need to calculate the number of moles.
The molar mass of gold (Au) is approximately 196.97 g/mol.
[tex]\[ \text{Number of moles} = \frac{9.85 \times 10^{-3} \, \text{g}}{196.97 \, \text{g/mol}} \approx 5.00085 \times 10^{-5} \, \text{mol} \][/tex]
Therefore, the number of moles represented by 9.85 × 10⁻³ g of gold is approximately [tex]\(5.00085 \times 10^{-5}\)[/tex] mol.
