Question

What mass of precipitate (in grams) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 57.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction: Na₃PO₄(aq) + Cr(NO₃)₃(aq) → CrPO₄(s) + 3NaNO₃(aq)?

Answer 1

The mass of the precipitate formed in the reaction of 45.5 mL of 0.300 M Na₃PO₄ with 57.0 mL of 0.200 M Cr(NO₃)₃ is approximately 1.812 grams, considering Cr(NO₃)₃ as the limiting reactant.

Step-by-step explanation:

To determine the mass of precipitate formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 57.0 mL of 0.200 M Cr(NO₃)₃, we must first identify the limiting reactant by comparing the number of moles of each reactant.

• Number of moles of Na₃PO₄ = Volume (L) × Molarity (mol/L) = 0.0455 L × 0.300 mol/L = 0.01365 mol.
• Number of moles of Cr(NO₃)₃ = Volume (L) × Molarity (mol/L) = 0.0570 L × 0.200 mol/L = 0.0114 mol.

Since the reaction is a 1:1 ratio between Na₃PO₄ and Cr(NO₃)₃, and there are fewer moles of Cr(NO₃)₃, it is the limiting reactant. The amount of CrPO₄ precipitate formed will be equal to the amount of Cr(NO₃)₃ that reacts.

Number of moles of CrPO₄ = 0.0114 mol (equal to moles of limiting reactant). To find the mass:

Mass of CrPO₄ = moles × molar mass (The molar mass of CrPO₄ is approximately 158.97 g/mol)

Mass of CrPO₄ = 0.0114 mol × 158.97 g/mol ≈ 1.812 g

Therefore, the mass of precipitate formed is approximately 1.812 grams.