Question

5.00 g CaH2 reacted with 6.00 g H2O in the reaction below. Calculate the % yield of H2 if a mass of 0.20 grams was collected after the reaction. CaHz(s) + 2H20(1) - Ca(OH)2(aq) + 2H2(g) 30% 42% 60% 84% 240%

Answer 1

CaH (s) + 2H₂O(1) - (a(OH)₂ (aq) + 2 H₂ (9) Initial soog 6.oog ha molah mass of Cat₂ = 42 g/mol. no of moles of Cat₂ = 5.00g - 0.1190 Mel чая Јмо no of moles of water = 6.00g - 0.3333 mal 189/mol By stochiometry 2 mol of Hao react with I mol of ca H₂ So, 0.3333 mal of H₂0 will react with 1 x 0.3333 ral - do 1666 mol But we only have 0.1190 mol of Cate in reaction mixture. so limiting reagent = (a H₂ I mol of cata produced a 2 mol of H₂ o.1190 mol will procluced = 2x0.1190 mol H₂ - 0.238 mol H₂ amount of H2 prwoluced (y) = 0.238 x 29 = 0.476 9. O model mol. %. yield of H₂ = actual yield x 100 Theortical yield = 0.209 X100 = 42.0%. 0.4769 so b option is correct.