Metal hydrides react with water to form hydrogen gas and the metal hydroxide. Fo
ID: 726982 • Letter: M
Question
Metal hydrides react with water to form hydrogen gas and the metal hydroxide. For example, SrH2(s) + 2H2O(l) => Sr(OH)2(s) + 2H2(g) What is the maximum mass of hydrogen gas that can be prepared from 50.0 g SrH2 and 48.0 g H2O?Explanation / Answer
SrH2(s) + 2H2O(l) =====> Sr(OH)2(s) + 2H2(g) Molar mass of SrH2 is = 87.62 + (2*1 ) = 89.62 g/mol Molar mass of H2O is = (2*1) + 16 = 18 g/mol According to the Equation , 89.62 g of SrH2 reacts with 2 * 18 = 36 g of H2O 50.0 g of SrH2 reacts with X g of H2O X = ( 50 * 36 ) / 89.62 = 20.08 g of H2O So ( 48 - 20.08 ) g of H2O left unreacted Since all the SrH2 consumed it is the limiting reactant Molar mass of H2 is = (2*1) = 2 g / mol According to the Equation , 89.62 g of SrH2 produces 2*2 = 4 g of H2 50 g of SrH2 produces Y g of H2 Y = ( 50 * 4 ) / 89.62 = 2.23 g of H2