Answer:
Mg
Explanation:
The standard reduction potentials are
<u>E°/V
</u>
Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42
Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85
Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Mg2+(aq) + 2e- ⟶ Mg(s); -2.38
The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
Mg is the only metal with a standard reduction potential lower than that of Cu, so
Only Mg will react spontaneously with Cu²⁺.
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:
0.045 L or 45 mL
Explanation:
Moles = Mass/M.Mass
Moles = 10 g / 109.94 g/mol
Moles = 0.09 moles
Also,
Molarity = Moles / Vol in L
Or,
Vol in L = Moles / Molarity
Vol in L = 0.09 mol / 2 mol/L
Vol in L = 0.045 L
145 since it’s the mass number plus protons I think
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
