Do you have a picture of where you go this question from?
Answer:
Longer hydrocarbon molecules have a stronger intermolecular force. More energy is needed to move them apart so they have higher boiling points . This makes them less volatile and therefore less flammable
Answer:
Theoretical yield = 2.5 g
Explanation:
Given data:
Mass of sodium = 79.7 g
Mass of water = 45.3 g
Theoretical yield of hydrogen gas = ?
Solution:
Chemical equation:
2Na + 2H₂O → 2NaOH + H₂
Number of moles of sodium:
Number of moles = mass/ molar mass
Number of moles = 79.7 g / 23 g/mol
Number of moles = 3.5 mol
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 45.3 g / 18g/mol
Number of moles = 2.5 mol
Now we will compare the moles of hydrogen gas with water and sodium.
H₂O : H₂
2 : 1
2.5 : 1/2×2.5 =1.25 mol
Na : H₂
2 : 1
3.5 : 1/2×3.5 =1.75 mol
water will be limiting reactant.
Theoretical yield:
Mass = number of moles × molar mass
Mass = 1.25 mol × 2 g/mol
Mass = 2.5 g
Answer:
Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>
- It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
- Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
- The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.
- <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>
∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.
- From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
- So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
- Finally, we can calculate the mass of copper produced using:
mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.
- <u><em>So, the answer is:</em></u>
<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>
