if you mixed 100 grams of Magnesium nitrite with 100 grams of Lithium Phosphide and made magnesium phosphide and lithium nitrite how many grams of each product would you make and how many grams of each reactant would be left over at the end?

This is a limiting reagent (LR) problem and you know that because amounts are given for both reactants.

3Mg(NO2)2 + 2Li3PO4 ==> Mg3(PO4)2 + 6LiNO2

mols Mg(NO2)2 = grams/molar mass = ?
mols Li3PO4 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Mg(NO2)2 to mols Mg3(PO4)2.
Do the same and convert mols Li3PO4 to mols Mg3(PO4)2.
It is likely that these two values for mols Mg3(PO4)2 will not agree; the correct value in LR problems is ALWAYS the smaller number and the reagent responsible for this is the LR. This identifies the LR. Remember this for later. That will be called the LR and the other reagent will be called OR (for other reagent).
Convert to grams by g = mols x molar mass.
Using the LR, convert, with the coefficients, mols LR to mols LiNO2. Then convert to grams by mols x molar mass. This takes care of all but the lat part of the problem; i.e., how much is left of each one.

How much is left of the LR. Answer is none since all of it is used up.
How to find mols OR used.
Using the coefficients in the balanced equation, convert mols of the LR used to mols OR used. Subtract initial mols - mols used = mols remaining.
Then convert to grams. g OR remaining = mols OR x molar mass OR.
Post all of your work if you get stuck.

To determine the grams of each product and the grams of each reactant left over at the end of the reaction, we need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.

First, we need to balance the equation for the reaction:
3Mg(NO2)2 + 2Li3P ⟶ Mg3P2 + 6LiNO2

The molar mass of Magnesium Nitrite (Mg(NO2)2) is:
Mg(NO2)2 = (24.31 g/mol + 2 * (14.01 g/mol + 16.00 g/mol)) = 148.31 g/mol

The molar mass of Lithium Phosphide (Li3P) is:
Li3P = (3 * (6.94 g/mol) + 30.97 g/mol) = 98.79 g/mol

Now, we can calculate the number of moles of each reactant:
Number of moles of Magnesium Nitrite = 100 g / 148.31 g/mol
Number of moles of Lithium Phosphide = 100 g / 98.79 g/mol

Next, we need to determine the stoichiometry of the reaction. From the balanced equation, we can see that the reaction ratio is 3:2 between Magnesium Nitrite and Lithium Phosphide.

Now, we compare the moles of Magnesium Nitrite and Lithium Phosphide to find the limiting reactant. The reactant with fewer moles is the limiting reactant.

This calculation will give us the number of moles of each reactant left over at the end of the reaction and the number of moles of each product formed.

Finally, we convert the moles of the products and the remaining reactants to grams, using their respective molar mass, to give the final answer in grams.

Note: Without the molar masses of Magnesium Phosphide and Lithium Nitrite, we cannot directly calculate the grams of each product formed and the grams of each reactant left over. We need to know the molar masses of those compounds beforehand.

If provided with the molar masses of Magnesium Phosphide (Mg3P2) and Lithium Nitrite (LiNO2), we can further calculate the grams of each product formed and the grams of each reactant left over.