The Equation is CaH2(s)+2H2O(l)-->Ca(OH)2(aq)+2H2(g) so,moles of H2 generated = PV / RT R = 62.363 L Torr K−1 mol−1 number of moles = 814 * 53.5 / 62.363 * 293 = 2.38 moles looking at the equation these came from 2.38 /2 moles of CaH2 or 1.19 moles Molar mass of CaH2 is 42.0943 g/mol so mass of CaH2 = 1.19 X 42.1 = 50.01g
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- An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Eigure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 4.40 to 2.20 L. When the external pressure is increased to 2.50 atm, the gas further compresses from 2.20 to 1.76 L. In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 4.40 to 1.76 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and g for the one-step process in joules? Express your answer with the appropriate units. View Available Hint(s) HA Value Submit Units Previous Answers ?