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Answer :
Answer:
We'll have 5.00 moles LiF and 2.50 moles Mg(NO3)2.
There will remain 2.50 moles MgF2
Explanation:
Step 1: data given
Number of moles lithium nitrate = 5.00 moles
Number of moles magnesium fluoride = 5.00 moles
Step 2: The balanced equation
2LiNO3 + MgF2 → 2LiF + Mg(NO3)2
Step 3: Calculate limiting reactant
For 2 moles LiNO3 we need 1 mol MgF2 to produce 2 moles LiF and 1 mol Mg(NO3)2
LiNO3 is the limiting reactant. It will completely be consumed (5.00 moles). MgF2 is in excess. There will react 2.50 moles . There will remain 5.00 - 2.50 = 2.50 moles MgF2.
Step 4: Calculate moles products
For 2 moles LiNO3 we need 1 mol MgF2 to produce 2 moles LiF and 1 mol Mg(NO3)2
For 5.00 moles LiNO3 we'll have 5.00 moles LiF and 2.50 moles Mg(NO3)2
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