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ks People Window Help 100% Nicole rn/takeAssignment/takeCovalentActivity.do?locator assignment gage Inbox-ricpather@...-(50,4 24 unread)-n. O GroupMe t My Print Center ® YouTube TV-Wate.. Use the References to access important values If meeded for this question. For the following reaction, 26.2 grams of sulfur dioxide are allowed to react with 5.65 grams of oxygen gas sulfur dioxide(g) +oxygen(g) sulfur trioxide(g) What is the maximum mass of sulfur trioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 9 more group attempts remaining Previous Show Hint

Explanation / Answer

Answer

Maximum mass of SO3 can be formed = 32.7g

Formula for the limiting reagent = SO2

Mass of excess reagent remains after reaction is complete = 2.38g

Explanation

2SO2(g) + O2(g) - - - - - - > 2SO3(g)

stoichiometrically, 2 moles of SO2 react with 1 mole of O2 to give 2 moles of SO3

No of moles of SO2 = 26.2g/64.07g/mol = 0.40893

No of moles of O2 =5.65g/16g/mol = 0.35313

0.35313 of moles O2 react with 0.70626 moles of SO2, but avoilable moles of SO2 is 0.40893 only

Therefore,

Limiting reagent is SO2

0.40893 moles of SO2 give 0.40893moles of SO3

Mass of SO3 maximum formed = 0.40893mol × 80.07g/mol =32.7g

No of moles of O2 remaining = 0.35313 - 0.20447= 0.14866

Mass of O2 remain unreacted = 0.14866mol× 16g/mol = 2.38g

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