Answer to Question #70602 in General Chemistry for steve

Question

What volume of 0.316 M Ba(OH)2 is required to react completely with 4.80 g sulfur?

Answer 1

3S + 3Ba(OH)2 = 2BaS + BaSO3 + 3 H2O
M(S)= 32.06 g/mol
From the reaction we see it takes 3 moles of sulfur to need to 3 mole of Ba(OH)2. Hence, from formula this is equivalent to 1:1 .
n=m/M=(4.80 g)/(32.06 g/mol)≅0.1497 mol
V=n/C=(0.1497 mol)/(0.316 mol/L)≅0.474 L
Answer
Need 0.474 L (0.316 M) Ba(OH)2 .