Find the limiting reagent and the reactant in excess when 100 mL of 0.2 mol/L Ba(OH)2 react completely with 50 mL of 0.5 mol/L of sulfuric acid. What is the mass of the precipitate.
2007-01-18 10:39:27 · 2 answers · asked by untilyoucamealong04 3 in Science & Mathematics ➔ Chemistry
0.1 L x 0.2 mol/L = 0.02 mol Ba(OH)2
0.050 L x 0.5 mol/l = 0.025 mol H2SO4
Ba(OH)2 + H2SO4 ==> BaSO4 + 2H2O
one mole one mole one mole 2 mole
Ba(OH)2 limits as you have only 0.02 mol
H2SO4 is in excess
Assuming complete reaction: 0.02 mole of BaSO4 will be formed
0.02 mol x 233g/ mol = 4.66 g BaSO4
2007-01-18 10:52:10 · answer #1 · answered by docrider28 4 · 0⤊ 0⤋
A restricting reagent is purely the reactant that runs out first for the duration of a reaction. As a reaction proceeds, reactants cut back, on an identical time as products boost. The restricting reagent is purely the reactant it truly is used up first. An extra reactant is purely the different reactant left over after the restricting reagent is used. there is mostly a restricting reagent, it purely relies upon on what number moles of what chemical you have.
2016-12-14 04:57:34 · answer #2 · answered by Anonymous · 0⤊ 0⤋