Metal hydrides react with water to form hydrogen gas and the metal hydroxide. Sr
ID: 736977 • Letter: M
Question
Metal hydrides react with water to form hydrogen gas and the metal hydroxide. SrH2(s) + 2 H2O(l) Sr(OH)2(s) + 2 H2(g) You wish to calculate the mass of hydrogen gas that can be prepared from 6.15 g of SrH2 and 5.21 g of H2O. (a) How many moles of H2 can be produced from the given mass of SrH2? mol,,,,,, (b) How many moles of H2 can be produced from the given mass of H2O? mol,,,,,,,, (c) Which is the limiting reactant? (Type your answer using the format CO2 for CO2.) (d) How many grams of H2 can be produced?Explanation / Answer
(a) 5.70g SrH2 x (1 mol SrH2 / 89.636g) (2 mol H2 / mol SrH2) (2.016g H2 / mol H2) = 0.256(3)g H2 (please note that the correct answer has three [3] significant figures, the final digit is to show the value used in the case of rounding.) (b) 4.75g H2O x (1 mol H2O / 18.016g H2O) (2 mol H2 / 2 mol H2O) (2.016g H2 / mol H2) = 0.532(5)g H2 (c) Because 4.75g of H2O can produce a larger quantity of H2 than 5.70g of SrH2, SrH2 is the limiting reactant. (d) Using the limiting reactant's value of 5.70g SrH2, 0.256g H2 could be produced 2. You need to write a balanced equation to solve this question: H2 + 2O2 -> 2H2O Convert g of hydrogen into mol H2: 0.0375g x (1 mol / 2.016g H2) = 0.018601 mol H2 H2 is in excess because there are more mol of H2 to begin with and it only requires 1 mol to react compared to 2 mol of O2 to react. To find how many g of H2O can form: 0.0185 mol O2 x ( 2 mol H2O / 2 mol O2) ( 18.016g H2O / mol H2O) = .333(2)g H2O How many grams of H2 remain: 0.0185 mol O2 x (1 mol H2 / 2 mol O2)(2.016g H2 / mol H2) = .018648g H2 Subtract that value from the initial measurement of H2 in g: .0375g - .018648 = 0.0189(5)g H2 remain after the reaction