Finding Limiting Reactants

Bananas

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Posts: 11

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11 years ago

I got this problem wrong. Can someone help me and tell me the steps to solving this correctly? Thank you!

What is the limiting reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2

3.4 mol Ca(NO3)2 x ( 6 mol LiNO3 / 3 mol Ca(NO3)2 ) x ( 69.95 g/mol LiNO3 / 1 mol LiNO3 ) = 475.66 g LiNO3

2.4 mol Li3PO4 x ( 6 mol LiNO3 / 2 mol Li3PO4 ) x ( 69.95 g/mol LiNO3/ 1 mol LiNO3 ) = 503.64 g LiNO3

Limiting Reactant = 475.66 g LiNO3

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Laser_3

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Subject Expert

11 years ago

Hi Bananas,

I think that you did everything right, but answering the question. The limiting reactant in this case would be Li3PO4 (the mass of LiNO3 just tell you how much of LiNO3 would be produced and help you determine which of the reactants is the limiting reactant).

Best,
Laser

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nyleve

wrote...

11 years ago

Hi there,

I think your set up is correct. However, there is no need to calculate the mass of LiNO3 if you are only need to find the limiting reactant. Therefore:

3.4mol Ca(NO3) x (6 LiNo3 / 3 Ca(NO3)) = 6.8 mol LiNO3
2.4mol Li3PO4 x (6 LiNO3 / 2 Li3PO4) = 7.2 mol LiNO3

Limiting reactant = Ca(NO3)2 .

~Nyleve

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LOL

Bananas Author

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11 years ago

I guess I over looked that one! Thank you!

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