Question

0.3 moles of BaCI2 is mixed with 0.2 moles of Na3PO4. Then the maximum number of moles of Ba3(PO4)2 that can be formed is

Answer 1

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Answer 2

Given...✍
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$3BaCl_{2}\: +\: 2Na_{3}PO_{4}$  ---> $Ba_{3}(PO_{4})_{2}\: + \:6NaCl$

0.3 mol of $3BaCl_{2}$ + 0.20 mol  of $2Na_{3}PO_{4}$ ---> x mol of $Ba_{3}(PO_{4})_{2}$

To find the maximum number of moles of $Ba_{3}(PO_{4})_{2}$... we have to find the limiting reagent...              

$\dfrac{0.3}{3}$ = $\dfrac{1}{10}$  of $3BaCl_{2}$  

$\dfrac{0.2}{2}$ = $\dfrac{1}{10}$ of $2Na_{3}PO_{4}$

On comparing what found is that both are same... no one is limiting reagant...

$\dfrac{1}{10}$ = $\dfrac{1}{10}$

Means.. question is wrong... the correct value of $3BaCl_{2}$ is 0.5 moles...

then it is solve like that...

$3BaCl_{2}\: +\: 2Na_{3}PO_{4}$  ---> $Ba_{3}(PO_{4})_{2}\: + \:6NaCl$

0.5 mol of $3BaCl_{2}$ + 0.20 mol  of $2Na_{3}PO_{4}$ ---> x mol of $Ba_{3}(PO_{4})_{2}$           

To find the maximum number of moles of $Ba_{3}(PO_{4})_{2}$... we have to find the limiting reagent...      
        
$\dfrac{0.5}{3}$ = $\dfrac{1}{6}$  of $3BaCl_{2}$  

$\dfrac{0.2}{2}$ = $\dfrac{1}{10}$ of $2Na_{3}PO_{4}$

On comparing we found is that...

$\dfrac{1}{6}$ > $\dfrac{1}{10}$

So, $Na_{3}PO_{4}$ is the limiting reagent....

Now,

$\dfrac{1}{10}$ = $\dfrac{x}{1}$

Cross multiply both of them with each other...

then we get,

10 x = 1

x = $\dfrac{1}{10}$

$\textbf{x = 0.1 mol}$

$Ba_{3}(PO_{4})_{2}$ is $\boxed{\boxed{\textbf{0.1 mol.}}}$