A piece of Calcium metal weighing 1.14 g is dropped into a beaker containing 55 mL of liquid water at 25 degrees Celsius. At this temp, the density of water is .997 g/mL. If the atmospheric pressure is 765 mm Hg, what volume in L of hydrogen gas will be released?

Question

anonymous · Answer

The equation is Ca + 2H2O --> Ca(OH)2 + H2. Sorry hehe. Please help!!!

anonymous · Answer

Ca + 2 H2O = Ca(OH)2 + H2 
(1.14 g Ca) / (40.0784 g Ca/mol) x (1 mol H2 / 1 mol Ca) = 0.028444 mol H2 
V = nRT / P = 
(0.028444 mol) x (62.36367 L mmHg/K mol) x (25 + 273)K / 
(765 mmHg) = 0.69 L H2

anonymous · Answer

Thank you!