Question

$H_{3}P{O}_4$ formed is converted into calcium phosphate by treating with $1$ mole of calcium hydroxide. Calcium phosphate obtained is :

• A. $2.5$
• B. $0.33$
• C. $0.1563$
• D. $1.25$

Answer 1

The correct option is B $0.33$

$310g{P}_4=\frac{310}{124}mol{P}_4=2.5mol{P}_4$
$2.5mol{P}_4{O}_{10}$ ($100\%$ yield)
$2.5\times \frac{50}{100}$ ($50\%$ yield)
$2.5\times 0.5\times 0.25\times 4$ ($25\%$ yield)
$=1.25mol{H}_{3}P{O}_4$
The balanced reaction is as follows:
$\underset{3mol}{3Ca(OH{)}_2}\underset{2mol}{+2H_{3}P{O}_4}⟶\underset{1mol}{C{a}_3(P{O}_4{)}_2}$
Given, $1$ mole $1.25$ mole -
So, $Ca(OH{)}_2$ is limiting reagent and the product formed is $1molCa(OH{)}_2,i.e.,\frac{1}{3}molC{a}_3(P{O}_4{)}_2=0.33moles.$