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Some Basic Concepts of Chemistry question

2023 · 6 Apr · Shift 1 · Q21
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Some Basic Concepts of Chemistry question

2023 · 6 Apr · Shift 1 · Q21

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
If 5 moles of BaCl2\mathrm{BaCl}_{2}BaCl2​ is mixed with 2 moles of Na3PO4\mathrm{Na}_{3} \mathrm{PO}_{4}Na3​PO4​, the maximum number of moles of Ba3(PO4)2\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}Ba3​(PO4​)2​ formed is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
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Correct answer: 1

  1. Write the balanced reaction

3 BaCl2+2 Na3PO4→Ba3(PO4)2+6 NaCl3\,\mathrm{BaCl_2} + 2\,\mathrm{Na_3PO_4} \rightarrow \mathrm{Ba_3(PO_4)_2} + 6\,\mathrm{NaCl}3BaCl2​+2Na3​PO4​→Ba3​(PO4​)2​+6NaCl

  1. Given moles
  • BaCl2=5\mathrm{BaCl_2} = 5BaCl2​=5 mol
  • Na3PO4=2\mathrm{Na_3PO_4} = 2Na3​PO4​=2 mol
  1. Use stoichiometric ratio to find limiting reagent

From the balanced equation:

3 mol BaCl2 react with 2 mol Na3PO43\text{ mol } \mathrm{BaCl_2} \text{ react with } 2\text{ mol } \mathrm{Na_3PO_4}3 mol BaCl2​ react with 2 mol Na3​PO4​

For 222 mol of Na3PO4\mathrm{Na_3PO_4}Na3​PO4​, required BaCl2\mathrm{BaCl_2}BaCl2​ is:

32×2=3 mol\frac{3}{2} \times 2 = 3\text{ mol}23​×2=3 mol

Available BaCl2=5\mathrm{BaCl_2} = 5BaCl2​=5 mol, so BaCl2\mathrm{BaCl_2}BaCl2​ is in excess.

Hence, Na3PO4\mathrm{Na_3PO_4}Na3​PO4​ is the limiting reagent.

  1. Calculate moles of product formed

From the equation:

2 mol Na3PO4→1 mol Ba3(PO4)22\text{ mol } \mathrm{Na_3PO_4} \rightarrow 1\text{ mol } \mathrm{Ba_3(PO_4)_2}2 mol Na3​PO4​→1 mol Ba3​(PO4​)2​

So, 222 mol of Na3PO4\mathrm{Na_3PO_4}Na3​PO4​ will form:

1 mol Ba3(PO4)21\text{ mol } \mathrm{Ba_3(PO_4)_2}1 mol Ba3​(PO4​)2​

  1. Nearest integer

1 mol1 \text{ mol}1 mol

So, the maximum number of moles of Ba3(PO4)2\mathrm{Ba_3(PO_4)_2}Ba3​(PO4​)2​ formed is:

1\boxed{1}1​

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