How many grams of solid barium sulfate form when 30.0 mL of 0.160 M barium chloride reacts with 70.0 mL of 0.065 M sodium sulfate? Aqueous sodium chloride forms also.

1.06g of BaSO₄ is produced

Explanation:

BaCl₂(aq) + Na₂SO₄(aq) ⇄ 2NaCl(aq) + BaSO₄(aq)

mole BaCl₂ = (0.16M × 0.030L) = 0.0048mole BaCl₂

mole Na₂SO₄ = (0.065M × 0.070L) = 0.00455mole Na₂SO₄

Amount of product BaSO₄ produce from each reagent

BaCl₂ = (0.0048 BaCl₂)  × ( 1 mol BaSO₄ / 1mol BaCl₂) × (233.3896g BaSO₄ / 1mol BaSO₄)

= 1.12g  BaSO₄

Na₂SO₄ = (0.00455 Na₂SO₄)  × ( 1 mol BaSO₄ / 1mol Na₂SO₄) × (233.3896g BaSO₄ / 1mol BaSO₄)

= 1.06g BaSO₄

i believe it is c.) 1575 j.