How many grams of barium sulfate, baso4, are produced if 25.34 ml of 0.113 m bacl2 completely react given the reaction: bacl2 + na2so4 → baso4 + 2 nacl?

Answer is: mass of barium sulfate is 0.668 grams.

Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.

V (BaCl₂) = 25.34 mL : 1000 mL/L = 0.02534 L.

c (BaCl₂) = 0.113 mol/L.

n (BaCl₂) = V (BaCl₂) · c (BaCl₂).

n (BaCl₂) = 0.02534 L · 0.113 mol/L.

n (BaCl₂) = 0.00286 mol.

From chemical reaction: n (BaCl₂) : n (BaSO₄) = 1 : 1.

n (BaSO₄) = 0.00286 mol.

m (BaSO₄) = n (BaSO₄) · M (BaSO₄).

m (BaSO₄) = 0.00286 mol · 233.4 g/mol.

m (BaSO₄) = 0.668 g.