A 500.0-mL sample of 0.200 M sodium phosphate is mixed with 400.0 mL of 0.289 M barium chloride. What is the mass of the solid produced?

Short Answer

Expert verified
The mass of the solid produced (barium phosphate) when a 500.0-mL sample of 0.200 M sodium phosphate is mixed with 400.0 mL of 0.289 M barium chloride is approximately 20.04 grams.

Step by step solution

01

Write the balanced chemical equation

The double displacement reaction between sodium phosphate (Na₃PO₄) and barium chloride (BaCl₂) will result in the formation of barium phosphate (Ba₃(PO₄)₂) and sodium chloride (NaCl). The balanced chemical equation is: 3 Na₃PO₄ (aq) + 2 BaCl₂ (aq) → Ba₃(PO₄)₂ (s) + 6 NaCl (aq) Step 2: Find the moles of each reactant
02

Find the moles of each reactant

First, we need to find the moles of each reactant. We can use the formula: moles = Molarity × Volume (in Liters) For sodium phosphate: moles = 0.200 M × 0.500 L = 0.100 moles For barium chloride: moles = 0.289 M × 0.400 L = 0.1156 moles Step 3: Determine the limiting reactant
03

Determine the limiting reactant

Compare the mole ratio of the reactants with the coefficients in the balanced chemical equation: Mole ratio (Na₃PO₄ : BaCl₂) = 0.100 moles : 0.1156 moles Divide the moles of each reactant by their respective coefficients: Na₃PO₄ = 0.100 moles / 3 = 0.0333 BaCl₂ = 0.1156 moles / 2 = 0.0578 The smallest value is for Na₃PO₄, so it is the limiting reactant. Step 4: Calculate the moles of solid product (Ba₃(PO₄)₂) formed
04

Calculate the moles of solid product formed

Based on the limiting reactant and its mole ratio in the balanced chemical equation, we can find the moles of barium phosphate formed: moles (Ba₃(PO₄)₂) = moles (Na₃PO₄, limiting reactant) × (1 mol Ba₃(PO₄)₂ / 3 mol Na₃PO₄) moles (Ba₃(PO₄)₂) = 0.100 moles × (1/3) = 0.0333 moles Step 5: Calculate the mass of the solid product (Ba₃(PO₄)₂) formed
05

Calculate the mass of the solid product formed

Using the molar mass of Ba₃(PO₄)₂, we can calculate its mass: Molar mass (Ba₃(PO₄)₂) = 3 × (Ba) + 2 × (3 × P + 4 × O) Molar mass (Ba₃(PO₄)₂) = 3 × (137.33 g/mol) + 2 × (3 × 30.97 g/mol + 4 × 16.00 g/mol) Molar mass (Ba₃(PO₄)₂) ≈ 601.93 g/mol mass (Ba₃(PO₄)₂) = moles (Ba₃(PO₄)₂) × Molar mass (Ba₃(PO₄)₂) mass (Ba₃(PO₄)₂) = 0.0333 moles × 601.93 g/mol ≈ 20.04 g Therefore, the mass of the solid produced (barium phosphate) is approximately 20.04 grams.

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