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In a solution, 1.50\,mole\,BaC{{l}_{2}} is added in 0.6\,mole\,N{{a}_{3}}P{{O}_{4}}, the maximum number of B{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}} that can be formed is?

Option: 1

0.5


Option: 2

0.1


Option: 3

0.3


Option: 4

1.0

 


Answers (1)

best_answer

Here is the balanced chemical reaction is:

3BaC{{l}_{2}}+2N{{a}_{3}}P{{O}_{4}}\to B{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}+6NaCl

According to reaction’s stoichiometry, 3 mole of BaC{{l}_{2}} combines with 2 mole of N{{a}_{3}}P{{O}_{4}}

\therefore 1.5\,mole\,BaC{{l}_{2}}  Reacts with 1\,mole\,N{{a}_{3}}P{{O}_{4}}, but we in reaction mixture sufficient amount of N{{a}_{3}}P{{O}_{4}} is not present, called limiting reagent. The amount of product will be formed according to the limiting reagent.

  \because 2\,mole\,N{{a}_{3}}P{{O}_{4}}\to 1\,mole\,B{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}} \\

\therefore 0.6\,mole\,N{{a}_{3}}P{{O}_{4}}\to \frac{1}{2}\times 0.6=0.3\,mole\,B{{a}_{3}}{{\left( P{{O}_{4}} \right)}_{2}} \\

 

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SANGALDEEP SINGH

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