Answer:
In none of the reactions ΔH°rxn equal to ΔH°f of the product.
Explanation:
The standard enthalpy of formation (ΔH°f) is the enthalpy change when 1 mole of a product is formed from its constituent elements in the standard states.
1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)
ΔH°rxn is NOT equal to ΔH°f of the product because H₂O(g) is not an element but a compound.
Na⁺(g) + F⁻(g) ⟶ NaF(s)
ΔH°rxn is NOT equal to ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).
K(g) + 1/2 Cl₂(g) ⟶ KCl(s)
ΔH°rxn is NOT equal to ΔH°f of the product because K is not in its standard state (K(s)).
O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)
ΔH°rxn is NOT equal to ΔH°f of the product because 2 moles of N₂O are formed.
In none of the above ΔHrxn equal to ΔHf of the product.
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>
Answer:
72.67g of B
Explanation:
The reaction of B₂O₃ to produce boron (B), is:
B₂O₃ → 3/2O₂ + 2B
<em>That means B₂O₃ produce 2 moles of boron</em>
Molar mass of B₂O₃ is 69.62g/mol. 234g of B₂O₃ contains:
234g B₂O₃ ₓ (1mol / 69.62g) = 3.361 moles of B₂O₃.
As 1 mole of B₂O₃ produce 2 moles of B, Moles of B that can be produced from B₂O₃ is:
3.361mol B₂O₃ ₓ 2 = <em>6.722 moles of B</em>.
As molar mass of B is 10.811g/mol. Thus mass of B that can be produced is:
6.722mol B ₓ (10.811g / mol) = <em>72.67g of B</em>
1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon,
then, (415.2/89.2) ×100 = 465.47 g of the substance will be required to yield 34.6 moles of carbon.
