ource from Cengage Learning Microsoft Edge ment/takeCovalentA catormassionment e

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ource from Cengage Learning Microsoft Edge ment/takeCovalentA catormassionment ent-take&takeAssignmentSessionLocator; assignment-take Use the References to access importast values if aeeded for this questios For the following reaction, 69.8 grams of barium hydroxide are allowed to react with 3S.0 grams of sulfuric acid barium hydroxidelag) + sulfuric acid(aq) barium sulfate(s) + water) What is the maximum amount of barium sulfate that can be formed? rams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Try Another Version 10 item attempts remaining

Explanation / Answer

Ba(OH)2 + H2SO4 -------> BaSO4 + 2H2O

according to balanced reaction

171.34 g Ba(OH)2 reacts with 98 g H2SO4  

69.8 g Ba(OH)2 reacts with 69.8 x 98 / 171.34 = 40.0 g H2SO4

but we have 38.0 g H2SO4. means H2SO4 is limiting reagent .

Limting reagent = H2SO4

now

98 g H2SO4 gives 233.38 g BaSO4

38 g H2SO4 gives 38 x 233.38 / 98 = 90.49 g BaSO4

mass of BaSO4 formed = 90.49 g

calculating exess reagent

98 g H2SO4 reacts with 171.34 g Ba(OH)2  

38 g H2SO4 reacts with 38 x 171.34 / 98 = 66.44 g Ba(OH)2

exess Ba(OH)2 left = 69.8 - 66.44 = 3.36 g

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