What is the percent yield of a reaction in which 84.4 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 8.44 mL of water (d = 1.00 g/mL)?
WO3 (s) + 3 H2 (g) --------------> W (s) + 3 H2O (l)
231.84 g 54.045 g
84.4 g ??
231.84 g WO3 ------------> 54.045 g H2O
84.4 g WO3 -------------> ??
mass of H2O produced = 84.4 x 54.045 / 231.84
= 19.7 g
actual mass of water = 8.44 x 1 = 8.44 g
percent yield = (actual / theoretical) x 100
= (8.44 / 19.7) x 100
percent yield = 42.9 %
What is the percent yield of a reaction in which 84.4 g of tungsten(VI) oxide (WO3)...