Question

Se (s) + 3Cl₂ (g) + 4H₂O (l) H₂SeO₄ (aq) + 6HCl (aq) Determine the limiting reactant in a mixture containing 146 g of Se, 464 g of Cl₂, and 113 g of H₂O. Calculate the maximum mass (in grams) of selenic acid, H₂SeO₄, that can be produced in the reaction.

Answer 1

The limiting reactant in the reaction is water (H₂O). Water's smallest mole to coefficient ratio indicates it will be completely consumed first, limiting the maximum amount of selenic acid (H₂SeO₄) that can be produced to 1223.107 grams.

Step-by-step explanation:

To determine the limiting reactant in the reaction between selenium (Se), chlorine (Cl₂), and water (H₂O), we first calculate the number of moles of each reactant using their respective molar masses from the periodic table: Se (78.96 g/mol), Cl₂ (70.90 g/mol), and H₂O (18.015 g/mol).

We find:

• Moles of Se = 146 g / 78.96 g/mol = 1.848 mol
• Moles of Cl₂ = 464 g / 70.90 g/mol = 6.543 mol
• Moles of H₂O = 113 g / 18.015 g/mol = 6.273 mol

Next, we divide the moles of each reactant by their coefficients in the balanced chemical reaction:

• Se : 1.848 mol / 1
• Cl₂ : 6.543 mol / 3 = 2.181 mol
• H₂O : 6.273 mol / 4 = 1.568 mol

H₂O has the smallest mole to coefficient ratio and is therefore the limiting reactant. Now, we calculate the maximum number of moles of H₂SeO₄ that can be formed from the limiting reactant:

Moles of H₂SeO₄ = Moles of H₂O = 6.273 mol

Lastly, using the molar mass of H₂SeO₄ (194.96 g/mol), we find the maximum mass of H₂SeO₄ produced:

Max mass of H₂SeO₄ = 6.273 mol * 194.96 g/mol = 1223.107 g