Metal hydrides react with water to from hydrogen gas and themetal hydroxide. SrH2(s) + 2H2O(l) Sr(OH)2(s) + 2 H2(g) You wish to calculate the mass of hydrogen gas that can beprepared from 5.19 g of SrH2 and 4.53 g ofH2O. A) How many moles of H2 can be produced from thegiven mass of SrH2? B) How many moles of H2 can be produced from thegiven mass of H2O? C) Which is the limiting reactant? D) How many grams of H2 can beproduced? Metal hydrides react with water to from hydrogen gas and themetal hydroxide. SrH2(s) + 2H2O(l) Sr(OH)2(s) + 2 H2(g) You wish to calculate the mass of hydrogen gas that can beprepared from 5.19 g of SrH2 and 4.53 g ofH2O. A) How many moles of H2 can be produced from thegiven mass of SrH2? B) How many moles of H2 can be produced from thegiven mass of H2O? C) Which is the limiting reactant? D) How many grams of H2 can beproduced?
Explanation / Answer
SrH2(s) + 2H2O(l) Sr(OH)2(s) + 2 H2(g) Molar mass of SrH2 = 87.6 + 2 * 1 = 89.6 g Molar mass of H2O = 2 * 1 + 16 = 18 g Molar mass of Sr(OH)2 = 87.6 + 2 ( 16 + 1 ) = 121.6g 89.6 g of SrH2 reacts with 2 * 18 g of H2O 5.19 g of SrH2 reacts with Xg of H2O X = ( 2*18*5.19)/ 89.6 = 2.085 g of H2O So , 4.53 - 2.085 = 2.445 g of H2O left unreacted 89.6 g of SrH2 produces 2 * 2 g of H2 5.19 g of SrH2 produces Y g of H2 Y = ( 2*2*5.19 ) / 89.6 = 0.2317 g of H2 A) How many moles of H2 can be produced from thegiven mass of SrH2? 89.6 g of SrH2 produces 2 moles of H2 5.19 g of SrH2 produces Z moles of H2 Z = ( 2*5.19) / 89.6 = 0.1158 moles of H2 B) How many moles of H2 can be produced from thegiven mass of H2O? what ever may be the mass of H2O , the no . of moles of H2formed is 0.1158 moles ( Since H2O is the excess reactant ) C) Which is the limiting reactant? The limiting reactant is SrH2 D) How many grams of H2 can be produced?----0.2317 g of H2