Cengage .. owLx2 | Online teaching and lG Chegg Study I Guided Solutio/X C cvg.c

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Cengage .. owLx2 | Online teaching and lG Chegg Study I Guided Solutio/X C cvg.cengagenow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator assignment-take&takeAssi; or assignment-take Visualize Empirical and Molecular F pts Use the References to access important values if needed for this question. For the following reaction, 24.6 grams of sulfur dioxide are allowed to react with 5.57 grams of water sulfur dioxide (g)+ water I)sulfurous acid (H2S03) (a) What is the maximum amount of sulfurous acid (H2SO3) that can be formed? 4. Chemicall Equations: Gram/Gra.. 1 pts 2req grams 5, Limiting Reactants: Compare Pr… 1 pts 2req What is the FORMULA for the limiting reagent? Question Question What amount of the excess reagent remains after the reaction is complete? Submit Answer Retry Entire Group 9 more group attempts remaining Progress: 3r5 groups Due Novw 30 at 11:55 PM Previous Next Finish Assignment Email Instructor Save and Exit Cengage Learning Cengage Technical Support

Explanation / Answer

Ans 1 :

Number of moles of SO2 = 24.6 / 64.06 = 0.384 moles

Number of moles of water = 5.57 / 18.0158 = 0.309 moles

Since equimolar concentrations of both the reactants are required , water is the limiting reagent

Formula - H2O

Number of moles of H2SO3 formed = 0.309 moles

Mass = 0.309 x 82.08 = 25.38 grams

Mass of SO2 utilised = 0.309 x 64.06 = 19.8 grams

Amount of excess SO2 remaining = 24.6 - 19.8

= 4.8 grams

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