Respuesta :
Answer:
To determine the amount of chromium chlorate produced and identify the limiting reactant, we first need to write and balance the chemical equation for the reaction between chromium chloride (CrCl₃) and oxygen (O₂) to form chromium chlorate (Cr(ClO₃)₃).
The balanced chemical equation is:
2 CrCl₃ + 3 O₂ → 2 Cr(ClO₃)₃
Now, let's determine the limiting reactant:
Calculate the moles of oxygen (O₂):
Moles of O₂ = volume of O₂ (in liters) / molar volume of O₂ at STP (22.4 L/mol)
= 12.5 L / 22.4 L/mol
≈ 0.558 moles
Calculate the mole ratio of CrCl₃ to O₂ from the balanced equation:
From the equation, 2 moles of CrCl₃ react with 3 moles of O₂.
Calculate the moles of CrCl₃:
Moles of CrCl₃ = 18 moles (given)
Determine the limiting reactant:
Calculate the moles of CrCl₃ required to react with the moles of O₂:
Moles of CrCl₃ required = (3/2) × moles of O₂
= (3/2) × 0.558 moles
≈ 0.837 moles
Since we have 18 moles of CrCl₃ available, which is much greater than 0.837 moles required, chromium chloride (CrCl₃) is in excess.
Now, let's calculate the moles of chromium chlorate (Cr(ClO₃)₃) produced using the limiting reactant (oxygen):
From the balanced equation, 3 moles of O₂ produce 2 moles of Cr(ClO₃)₃.
Moles of Cr(ClO₃)₃ produced = (2/3) × moles of O₂
= (2/3) × 0.558 moles
≈ 0.372 moles
Finally, let's convert moles of chromium chlorate to grams using its molar mass:
Molar mass of Cr(ClO₃)₃ = (1 * atomic mass of Cr) + (3 * (atomic mass of Cl + atomic mass of O))
= (1 * 51.9961 g/mol) + (3 * (35.453 g/mol + 15.999 g/mol))
≈ 266.357 g/mol
Mass of chromium chlorate produced = moles of Cr(ClO₃)₃ × molar mass of Cr(ClO₃)₃
≈ 0.372 moles × 266.357 g/mol
≈ 98.9 grams
So, approximately 98.9 grams of chromium chlorate will be produced. And oxygen is the limiting reactant
Explanation:
