Question

if 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4 find the maximum no of moles of Ba3(PO4)2 formed.

Answer 1

since sodium phosphate is the limiting reagent, 0.2 moles of it will yield 0.1 mol of barium phosphate.

Answer 2

Answer:

$0.1mol$ of $Ba_{3}(PO_{4})_{2}$ will be produced.

Explanation:

$3BaCl_{2}+ 2Na_{3}PO_{4}$ → $Ba_{3}(PO_{4})_{2} + 6NaCl$

Three moles of $BaCl_{2}$ react with two moles of $Na_{3}PO_{4}$ to produce one mole of $Ba_{3}(PO_{4})_{2}$ and six moles of $NaCl$.

We are asked to find the maximum number of moles of $Ba_{3}(PO_{4})_{2}$ produced when 0.5 moles of $BaCl_{2}$ and 0.2 moles of $Na_{3}PO_{4}$ react.

First, we find the limiting reagent by dividing the moles of each reactant by their coefficient.

$BaCl_{2}$ - $\frac{0.5}{3}= 0.167$

$Na_{3}PO_{4}$- $\frac{0.2}{2}=0.1$

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

Now,

2 moles of $Na_{3}PO_{4}$ → 1 mole of $Ba_{3}(PO_{4})_{2}$

0.2 moles of $Na_{3}PO_{4}$ → $x$ moles of $Ba_{3}(PO_{4})_{2}$

$1 * 0.2 = 2 * x$

$\frac{0.2}{2} = 0.1 mol$ of $Ba_{3}(PO_{4})_{2}$