owtv2 Onlne teaching com/ilmytakeAssignment/takeCovalentActivity.do?locator assi
ID: 1025899 • Letter: O
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owtv2 Onlne teaching com/ilmytakeAssignment/takeCovalentActivity.do?locator assignment-take&takeAssignmentSessionL.ocator; assignment-take Use the Refereaces to access inmportant values if needed for this question For the following reaction, 128 grams of sulfur dioxide are allowed to react with 96.0 grams of oxygen gas What is the FORMULA for the limiting reagent? What is the maximom amount of sulfur trioxide that can be formed? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Iry Another Version item attempts remaining e to search o e F12 PrtScrInsert 6 8 9 Backs 0Explanation / Answer
2 SO2 (g) + O2 (g) ------> 2 SO3 (g)
number of moles of SO2 = 128g/64g/mol = 2 moles
number of moles of O2 = 96g/32g/mol = 3 moles
2 moles of SO2 requires 1 mole of O2 according to equation then
So2 is limiting rreagent and O2 is excess reagent.
2 moles of SO2 produces 2 moles of SO3 then,
mass of SO3 formed = 2 * 80 = 160 g
remaining excess reagent = 3 - 1 = 2 moles
mass of O2 remaining = 2 mol * 32 g/mol = 64 g
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