Calcium hydride (cah2) reacts with water to form hydrogen gas: cah2(s) + 2h2o(l) → ca(oh)2(aq) + 2h2(g) how many grams of cah2 are needed to generate 58.0 l of h2 gas at a pressure of 0.811 atm and a temperature of 32°c?
Using the ideal gas law equation, we can find the number of H₂ moles produced. PV = nRT Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa V - volume - 58.0 x 10⁻³ m³ R - universal gas constant - 8.314 Jmol⁻¹K⁻¹ T - temperature - 32 °C + 273 = 305 K substituting these values in the equation, 82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K n = 1.88 mol
The balanced equation for the reaction is as follows; CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g) stoichiometry of CaH₂ to H₂ is 1:2 When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol therefore number of CaH₂ moles reacted = 0.94 mol Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed
