Answer:
when the forward and reverse reactions occur at equal rates.
chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.
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>
oxygen is in gaseous state at this temperature
Answer:
Option b. 0.048 M
Explanation:
We have the molecular weight and the mass, from sulcralfate.
Let's convert the mass in g, to moles
1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.
Molarity is mol /L
Let's convert the volume of solution in L
10 mL . 1L/1000 mL = 0.01 L
4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L
