Points out of 1.00 Not flagged Question text Suppose 5.00 moles of Lithium Nitrate react with 5.00 moles of Magnesium Fluoride react according to BALANCE 1st!!: __Li(NO3) __ MgF2 -> __LiF __Mg(NO3)2

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

the objects have the same charge.