Given the reaction 2H2O2 → O2 + 2H2O, how many grams of water are produced if 20.0g of Hydrogen Peroxide decompose? 2. Corrected_text: Given the reaction 3Fe + 4H2O → Fe3O4 + 4H2, how many grams of water are needed to react with 42.0g of Iron? 3. Corrected_text: Given the reaction 2Al + 6HCl → 2AlCl3 + 3H2, how many grams of Aluminum Chloride are produced if 4.80 moles of HCl are consumed
The Correct Answer and Explanation is:
Let’s break down each problem step by step:
1. Hydrogen Peroxide Decomposition Reaction:
Given reaction:
2H2O2→O2+2H2O2H_2O_2 \rightarrow O_2 + 2H_2O2H2O2→O2+2H2O
Given: 20.0 g of hydrogen peroxide (H₂O₂)
We need to calculate how many grams of water (H₂O) are produced.
Step 1: Calculate the moles of H₂O₂
The molar mass of H₂O₂ is calculated as:
2×(1.008 g/mol (for H))+2×(15.999 g/mol (for O))=34.014 g/mol2 \times (1.008 \, \text{g/mol} \, \text{(for H)}) + 2 \times (15.999 \, \text{g/mol} \, \text{(for O)}) = 34.014 \, \text{g/mol}2×(1.008g/mol(for H))+2×(15.999g/mol(for O))=34.014g/mol
Now, calculate the moles of H₂O₂ in 20.0 g: moles of H2O2=20.0 g34.014 g/mol≈0.588 mol\text{moles of H}_2\text{O}_2 = \frac{20.0 \, \text{g}}{34.014 \, \text{g/mol}} \approx 0.588 \, \text{mol}moles of H2O2=34.014g/mol20.0g≈0.588mol
Step 2: Use stoichiometry to find moles of H₂O produced
From the balanced chemical equation, we see that 2 moles of H₂O₂ produce 2 moles of H₂O. Thus, the ratio is 1:1. Therefore, the moles of water produced are the same as the moles of H₂O₂.
So, 0.588 mol of H₂O₂ will produce 0.588 mol of H₂O.
Step 3: Convert moles of H₂O to grams
The molar mass of water (H₂O) is:
2×(1.008 g/mol)+1×(15.999 g/mol)=18.016 g/mol2 \times (1.008 \, \text{g/mol}) + 1 \times (15.999 \, \text{g/mol}) = 18.016 \, \text{g/mol}2×(1.008g/mol)+1×(15.999g/mol)=18.016g/mol
Now, calculate the mass of water produced: mass of H2O=0.588 mol×18.016 g/mol=10.6 g\text{mass of H}_2\text{O} = 0.588 \, \text{mol} \times 18.016 \, \text{g/mol} = 10.6 \, \text{g}mass of H2O=0.588mol×18.016g/mol=10.6g
So, 10.6 grams of water are produced.
2. Iron and Water Reaction:
Given reaction:
3Fe+4H2O→Fe3O4+4H23Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_23Fe+4H2O→Fe3O4+4H2
Given: 42.0 g of Iron (Fe)
We need to calculate how many grams of water (H₂O) are needed to react with 42.0 g of iron.
Step 1: Calculate the moles of iron (Fe)
The molar mass of iron (Fe) is:
55.845 g/mol55.845 \, \text{g/mol}55.845g/mol
Now, calculate the moles of Fe in 42.0 g: moles of Fe=42.0 g55.845 g/mol≈0.752 mol\text{moles of Fe} = \frac{42.0 \, \text{g}}{55.845 \, \text{g/mol}} \approx 0.752 \, \text{mol}moles of Fe=55.845g/mol42.0g≈0.752mol
Step 2: Use stoichiometry to find moles of H₂O needed
From the balanced chemical equation, we see that 3 moles of Fe react with 4 moles of H₂O. Thus, the ratio of Fe to H₂O is 3:4.
Now, calculate the moles of water needed: moles of H2O=43×0.752 mol Fe≈1.003 mol H2O\text{moles of H}_2\text{O} = \frac{4}{3} \times 0.752 \, \text{mol Fe} \approx 1.003 \, \text{mol H}_2\text{O}moles of H2O=34×0.752mol Fe≈1.003mol H2O
Step 3: Convert moles of H₂O to grams
The molar mass of water (H₂O) is:
18.016 g/mol18.016 \, \text{g/mol}18.016g/mol
Now, calculate the mass of water needed: mass of H2O=1.003 mol×18.016 g/mol≈18.1 g\text{mass of H}_2\text{O} = 1.003 \, \text{mol} \times 18.016 \, \text{g/mol} \approx 18.1 \, \text{g}mass of H2O=1.003mol×18.016g/mol≈18.1g
So, 18.1 grams of water are needed to react with 42.0 g of iron.
3. Aluminum Chloride Production:
Given reaction:
2Al+6HCl→2AlCl3+3H22Al + 6HCl \rightarrow 2AlCl_3 + 3H_22Al+6HCl→2AlCl3+3H2
Given: 4.80 moles of HCl
We need to calculate how many grams of Aluminum Chloride (AlCl₃) are produced.
Step 1: Use stoichiometry to find moles of AlCl₃ produced
From the balanced chemical equation, we see that 6 moles of HCl produce 2 moles of AlCl₃. Thus, the ratio of HCl to AlCl₃ is 6:2 or 3:1.
Now, calculate the moles of AlCl₃ produced: moles of AlCl3=26×4.80 mol HCl=1.60 mol AlCl3\text{moles of AlCl}_3 = \frac{2}{6} \times 4.80 \, \text{mol HCl} = 1.60 \, \text{mol AlCl}_3moles of AlCl3=62×4.80mol HCl=1.60mol AlCl3
Step 2: Convert moles of AlCl₃ to grams
The molar mass of AlCl₃ is:
26.98 g/mol (for Al)+3×35.45 g/mol (for Cl)=133.34 g/mol26.98 \, \text{g/mol} \, \text{(for Al)} + 3 \times 35.45 \, \text{g/mol} \, \text{(for Cl)} = 133.34 \, \text{g/mol}26.98g/mol(for Al)+3×35.45g/mol(for Cl)=133.34g/mol
Now, calculate the mass of AlCl₃ produced: mass of AlCl3=1.60 mol×133.34 g/mol=213.34 g\text{mass of AlCl}_3 = 1.60 \, \text{mol} \times 133.34 \, \text{g/mol} = 213.34 \, \text{g}mass of AlCl3=1.60mol×133.34g/mol=213.34g
So, 213.34 grams of Aluminum Chloride are produced when 4.80 moles of HCl are consumed.
Summary of Answers:
- Water produced: 10.6 grams
- Water needed: 18.1 grams
- Aluminum Chloride produced: 213.34 grams
