what is the mass of lithium sulfate when 45 grams of lithium reacts with excess silver sulfate

Answer 1

Write the balanced equation: 2Li(s) + Ag₂SO₄(aq) ==> Li₂SO₄(aq) + 2Ag(s)

moles of Li used = 45 g x 1 mole Li/6.9 g = 6.5 moles Li

moles Li₂SO₄ produced = 6.5 moles Li x 1 mole Li₂SO₄/2 moles Li = 3.26 moles Li₂SO₄ produced

mass Li₂SO₄ produced = 3.26 moles x 109.9 g/mole = 358 g = 360 g (to 2 significant figures)