(a) For the balanced equation shown below, if 10 grams of Ca(OH)2 reacted with 16.5 grams...

(a) For the balanced equation shown below, if 10 grams of Ca(OH)2 reacted with 16.5 grams H3PO4, How many grams of hydroxide (OH) woudl be produced? (Hint: assuming that the reactants reacted completely. You are asked to find the theoretical yield for the hydroxide)

3Ca(OH)2 + 2H3PO4 ------------> Ca3(PO4)2 + 6OH

(b) assuming that the actual yield is 3.085 grams from the reaction from (a), calculate the percent yield using your answer from (a).

Solutions

Expert Solution

given:

3Ca(OH)₂ + 2H₃PO₄ ------------> Ca₃(PO₄)₂ + 6OH

weight of Ca(OH)₂= 10g

weight of H₃PO₄= 16.5g

molecular weight of Ca(OH)₂= 74.093g/mol

molecular weight of H₃PO₄= 97.994g/mol

No of moles of Ca(OH)₂=10/74.093=0.1349 moles

No Of moles Of H₃PO₄= 16.5/97.994=0.1683 moles

when we are determining limiting reactant we have to divide no of moles of reactant present with stoichiometry coefficient in balanced reaction . in the balanced reaction Ca(OH)₂ stoichiomery coefficient is 3. and that of H₃PO₄ is 2. when we divide no of moles by stoichiometric coefficient for Ca(OH)₂ it comes out to be 0.04496 and for H₃PO₄ the value is 0.08415. comparing two values 0.04496 is less than 0.08415. so Ca(OH)₂ is limiting reactant.

now the amount of product will be determined by Ca(OH)₂.

when we see the balanced reaction the ratio between Ca(OH)₂ and OH^- is 1:2.

no of moles of Ca(OH)₂ =O.1349moles, so No of moles OH^-produced will be 0.1349*2=0.2698 moles .

weight of OH^- should be produced= 0.2698*17=4.586g

Actual yield observed= 3.085g

Percentage yield=(3.085/4.586)*100= 67.2699%

queen_honey_blossom answered 1 year ago

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