Answer to Question #184449 in General Chemistry for dustin

Question

1. What mass of barium sulfate (233.40 g/mol) will be made when 75.0 mL of 0.450M barium chloride reacts with 25.0 mL of  1.25M Sodium Sulfate? 

BaCl2 +  Na2SO4 → 2 NaCl + BaSO4

Answer 1

$BaCl_2 + Na_2SO_4 → 2 NaCl + BaSO_4$

Mole ratio is 1:1 this implies that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4

Moles n = Molarity C × Volume in liters V

Moles of BaCl2 = 0.075×0.45 = 0.034 moles

Moles of Na2SO4 = 1.25×0.025 =0.031 moles

Mass = Molar mass × moles

Mass of BaSO4 = (0.031+0.034)mol×233.4g/mol =15.171g

Ans = 15.17g