Metal hydrides react with water to form hydrogen gas and the metal hydroxide. Sr
ID: 736810 • 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 5.90 g of SrH2 and 5.05 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?
........ g
Explanation / Answer
one way is to answer questions a & b, which ever produces the least is the limiting reagent another way is: SrH2+2H2O --> find the grams that react: SrH2: 89.64 g/mole SrH2 2H2O: 2moles @ 18.02 g/mole = 36.04 grams H2O ratio: (89.64 g SrH2) / (36.04 g H2O) = 2.49 times as many grams of SrH2 are needed to react ,...as compared to grams of H2O then look @ what they dumped in: (7.02g SrH2) / (5.85g H2O) = 1.20 times as many grams of SrH2 were used.... that's not enough..... SrH2 is the limiting reagent ======================================… 1) How many moles of H2 can be produced from the given mass of SrH2, using the ratio 1SrH2 =>2H2 & molar mass: 7.02g SrH2 @ 2 mol H2 / 89.64 g/mole SrH2 = 0.1566 moles of H2 your answer(3sigfigs): 0.157 moles H2 ========================== b)How many moles of H2 can be produced from the given mass of H2O, using the ratio 2H2O => 2H2 & molar mass: 5.85g H2O @ 2 mol H2 / (2mol H2O)(18.02 g/mole) = 0.3246 mol H2 your answer(3sigfigs): 0.325 moles H2 ====================================== c) Which is the limiting reagent, SrH2 or H2O? SrH2 because it produces less, =================================== d) How many grams of H2 can be produced? 0.1566 moles H2 @ 2.02 g/mol = 0.31633 grams of H2 your answer(3sigfigs): 0.316 g H2