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Chapter 4: 4.8 CYL (page 194)

How many moles of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce Ca3(PO4)2 according to the equation \(3Ca{\left( {OH} \right)_2} + 2{H_3}P{O_4} \to C{a_3}{\left( {P{O_4}} \right)_2} + 6{H_2}O\)?

Short Answer

Expert verified

2.04 mol of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce 1.36 mol of Ca3(PO4)2.

Step by step solution

01

Given data

From the given balanced equation, we can see that 3 moles of Ca(OH)2 reacted with 2 moles of H3PO4.

Therefore, let the number of moles be X.

X moles of Ca(OH)2 react with 1.36 moles of H3PO4.

X=\(\frac{{3 \times 1.36}}{2}\)

X = moles of Ca(OH)2= 2.04 mol

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