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Q. The maximum number of moles of $\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2$ that can be formed if $2mole$ $BaCl_{2}$ is mixed with $1mole$ $Na_{3}PO_{4}$ is

NTA AbhyasNTA Abhyas 2020Some Basic Concepts of Chemistry

Solution:

The balanced chemical reaction is

$3 \mathrm{BaCl}_2+2 \mathrm{Na}_3 \mathrm{PO}_4 \rightarrow \mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2+6 \mathrm{NaCl}$

According to the reaction,

3 mole $\mathrm{BaCl}_2$ reacts with 2 mole $\mathrm{Na}_3\left(\mathrm{PO}_4\right)_2$

Thus, $2moleBaCl_{2}$ will require $\frac{2}{3}\times 2=\frac{4}{3}=1.33moleNa_{3}PO_{4}$

Since sodium phosphate is taken in less quantity then the required quantity, here, $Na_{3}PO_{4}$ is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent.

According to the reaction.

2 mole $Na _{3} PO _{4}$ gives 1 mole $Ba _{3}\left( PO _{4}\right)_{2}$

So, 1 mole $Na _{3} PO _{4}$ will give $0.5$ mole $Ba _{3}\left( PO _{4}\right)_{2}$.