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If 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4 ,then maximum number of moles of Ba3(PO4)2 that can be formed
- a)0.5
- b)0.1
- c)0.3
- d)0.7
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If 0.5 mole of BaCl2is mixed with 0.2 mole of Na3PO4,then maximum numb...
Balanced equation for reaction between BaCl2 and Na3PO4 is as follows:
3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl
3 moles of BaCl2 react with 2 moles of Na3PO4 to give 1 mole of Ba3(PO4)2
0.5 moles of BaCl2 will react with (2/3) x 0.5 = 0.33 moles of Na3PO4
Available moles of Na3PO4 = 0.2
So, Na3PO4 is the limiting reagent
Now, 2 moles of Na3PO4 give 1 mole of Ba3(PO4)2
So, 0.2 moles of Na3PO4 will give (1/2) x 0.2 = 0.1 mole of Ba3(PO4)2
Hence, maximum number of moles of Ba3(PO4)2 formed = 0.1
Most Upvoted Answer
If 0.5 mole of BaCl2is mixed with 0.2 mole of Na3PO4,then maximum numb...
Given:
- Moles of BaCl2 = 0.5
- Moles of Na3PO4 = 0.2
To find:
- Maximum number of moles of Ba3(PO4)2 that can be formed
Explanation:
The balanced chemical equation for the reaction between BaCl2 and Na3PO4 is:
3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl
From the balanced equation, we can see that 3 moles of BaCl2 react with 2 moles of Na3PO4 to form 1 mole of Ba3(PO4)2.
Step 1: Determine the limiting reagent
To find the limiting reagent, we need to compare the moles of each reactant with the stoichiometric ratio in the balanced equation.
For BaCl2:
Moles of BaCl2 = 0.5
Moles of Ba3(PO4)2 formed from 0.5 moles of BaCl2 = 0.5/3 = 0.1667
For Na3PO4:
Moles of Na3PO4 = 0.2
Moles of Ba3(PO4)2 formed from 0.2 moles of Na3PO4 = 0.2/2 = 0.1
The limiting reagent is the one that produces the lesser amount of product. In this case, Na3PO4 is the limiting reagent as it produces 0.1 moles of Ba3(PO4)2, which is less than what BaCl2 produces.
Step 2: Calculate the maximum moles of Ba3(PO4)2 formed
Since Na3PO4 is the limiting reagent, the maximum moles of Ba3(PO4)2 that can be formed is 0.1.
Answer:
The correct option is B) 0.1 moles.
- Moles of BaCl2 = 0.5
- Moles of Na3PO4 = 0.2
To find:
- Maximum number of moles of Ba3(PO4)2 that can be formed
Explanation:
The balanced chemical equation for the reaction between BaCl2 and Na3PO4 is:
3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl
From the balanced equation, we can see that 3 moles of BaCl2 react with 2 moles of Na3PO4 to form 1 mole of Ba3(PO4)2.
Step 1: Determine the limiting reagent
To find the limiting reagent, we need to compare the moles of each reactant with the stoichiometric ratio in the balanced equation.
For BaCl2:
Moles of BaCl2 = 0.5
Moles of Ba3(PO4)2 formed from 0.5 moles of BaCl2 = 0.5/3 = 0.1667
For Na3PO4:
Moles of Na3PO4 = 0.2
Moles of Ba3(PO4)2 formed from 0.2 moles of Na3PO4 = 0.2/2 = 0.1
The limiting reagent is the one that produces the lesser amount of product. In this case, Na3PO4 is the limiting reagent as it produces 0.1 moles of Ba3(PO4)2, which is less than what BaCl2 produces.
Step 2: Calculate the maximum moles of Ba3(PO4)2 formed
Since Na3PO4 is the limiting reagent, the maximum moles of Ba3(PO4)2 that can be formed is 0.1.
Answer:
The correct option is B) 0.1 moles.
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Community Answer
If 0.5 mole of BaCl2is mixed with 0.2 mole of Na3PO4,then maximum numb...
According To the question Na3PO4 os limiting reagent and as you know limiting is deciding . After balanceing the equation ,
2 moles of Na3PO4 gives 1 mole of Ba3 (PO4)2 0.2 moles of Na3PO4 gives 0.1 mole os Ba3 (PO4)2
2 moles of Na3PO4 gives 1 mole of Ba3 (PO4)2 0.2 moles of Na3PO4 gives 0.1 mole os Ba3 (PO4)2
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If 0.5 mole of BaCl2is mixed with 0.2 mole of Na3PO4,then maximum number of moles of Ba3(PO4)2that can be formeda)0.5b)0.1c)0.3d)0.7Correct answer is option 'B'. Can you explain this answer?
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