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now.com/ilrn/takeAssignment/takeCovalentActivity.do?locator-assignment-take&takeAssignmentSessionLocator-as; Use the References to access important values if needed for this question For the following reaction, 14.7 grams of carbon dioxide are allowed to react with 41.8 grams of potassium hydroxide carbon dioxide(g)+ potassium hydroxide(aq) potassium carbonate(aq)+ water) What is the maximum amount of potassium carbonate that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining

Explanation / Answer

CO2(g) + 2KOH(aq) = K2CO3(aq) + H2O(l)

No. Of moles of CO2 = Mass/molar mass = 14.7g/44g/mole = 0.334 mole .

No. Of moles of KOH = 41.8g/56.11g/mole = 0.745 .

As one mole of CO2 reacts will 2 moles of KOH , and thus 0.334 mole of CO2 will react with = 0.334 ×2 = 0.668 mole of KOH and we have 0.745 mole of KOH , Thus no. Of moles of KOH are extra .

And hence CO2 is the limiting reagent .

Amount of potassium carbonate formed depends on the amount of limiting reagent .

Thus as 1 mole of CO2 gives 1 mole of potassium carbonate ,

Then 0.334 mole of CO2 will produce = 0.334 mole of potassium carbonate .

Mass of potassium carbonate = mole × molar mass = 0.334 × 138.205 = 46.16 g of potassium carbonate will be produced .

Now , extra mole of KOH = 0.745 - 0.668 = 0.077 mole .

Mass of extra KOH after reaction = 0.077 × 56.11 = 4.32 g

Formula of limiting reagent is CO2 .

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