if you begin with 995g of CH4 and 2510g of water:CH 4 + H 2 O ---> CO +3H 2 a) w
ID: 76883 • Letter: I
Question
if you begin with 995g of CH4 and 2510g of water:CH4 + H2O ---> CO +3H2 a) what is the maximum mass of hydrogen gas that can beproduced? b) If only 200 g of hydrogen gas are actually isolated, whatis the percent yield? if you begin with 995g of CH4 and 2510g of water:CH4 + H2O ---> CO +3H2 a) what is the maximum mass of hydrogen gas that can beproduced? b) If only 200 g of hydrogen gas are actually isolated, whatis the percent yield? a) what is the maximum mass of hydrogen gas that can beproduced? b) If only 200 g of hydrogen gas are actually isolated, whatis the percent yield?Explanation / Answer
995 g CH4 *(1 mol CH4/ 16 g CH4) * (1 mol H2O/1 mol CH4) = 62.1875moles of H2O required to react with 995 g CH4 2510 g H2O * (1mol H2O/18 g) = 139.444444 moles H2O available. So CH4 is limiting reactant 995 g CH4 *(1 mol CH4/ 16 g) *(3 mol H2 / 1 mol CH4) *(2 g H2/ 1mol H2) = 373.12500 g H2 ~ 373 g H2 b) % yield = 200 g / 373.125 g = 0.5360134 ~ 53.6 %
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