1)For the following reaction, 17.4 grams of carbon dioxide are allowed to react
ID: 1016416 • Letter: 1
Question
1)For the following reaction, 17.4 grams of carbon dioxide are allowed to react with 49.2 grams of potassium hydroxide . carbon dioxide(g) + potassium hydroxide(aq) potassium carbonate(aq) + water(l)
-What is the maximum amount of potassium carbonate that can be formed? ____grams
-What is the FORMULA for the limiting reagent?
-What amount of the excess reagent remains after the reaction is complete? ______grams
2)
For the following reaction, 4.35 grams of diphosphorus pentoxide are mixed with excess water . The reaction yields 4.73 grams of phosphoric acid .
What is the theoretical yield of phosphoric acid ? grams What is the percent yield for this reaction ? %Explanation / Answer
2KOH + CO2 -------> K2CO3 + H2O
no of moles of Co2 = W/G.M.Wt = 17.4/44 = 0.395 moles
no of moles of KOH = W/G.M.Wt = 49.2/56 = 0.878moles
1 mole of Co2 react with 2 moles of KOH
0.395 moles of CO2 react with = 2*0.395/1 = 0.79 moles of KOH
limiting reagent is CO2
Excess reagent is KOH
1 mole of CO2 react with KOH to form 1 mole of K2CO3
0.395 moles of CO2 react with KOH to form = 0.395 moles of K2CO3
mass of K2CO3 = no of moles * gram molar mass
= 0.395*138 = 54.51g of K2CO3
no of moles of excess reagent left over = 0.878-0.79 = 0.088 moles of KOH
mass ofexcess reagent left over KOH = 0.088*56 = 4.928g of KOH
2. P2O5 + 3H2O ----->2H3PO4
1 mole of P2O5 react with H2O to form 2 moles of H3PO4
142 g of P2O5 react with H2O to form 2*98g of H3PO4
4.35g of P2O5 react with H2O to form = 2*98*4.35/142 = 6g of H3PO4
theoretical yield = 6g
percentage of yield = actual yield*100/theoretical yield
= 4.73*100/6 = 78.83% >>>> answer
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