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.
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To determine the amount of each product and the amount of reactant remaining, we need to find the limiting reactant in the given chemical reaction. The limiting reactant is the one that is completely consumed, thus determining the maximum amount of product that can be formed.
Let's start by writing the balanced chemical equation for the reaction:
3Mg(NO3)2 + 2Li3P -> 2Mg3P2 + 6LiNO3
From the balanced equation, we can see that the molar ratio of Mg(NO3)2 to Mg3P2 is 3:2, and the molar ratio of Li3P to Mg3P2 is 2:1.
Step 1: Calculate the number of moles of each reactant.
- Mg(NO3)2 (Magnesium Nitrite): = (mass of Mg(NO3)2) / (molar mass of Mg(NO3)2)
= 100 g / (148.31 g/mol)
= 0.674 mol
- Li3P (Lithium Phosphide): = (mass of Li3P) / (molar mass of Li3P)
= 100 g / (115.79 g/mol)
= 0.864 mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, we need to compare the mole ratio of the reactants to their coefficients in the balanced equation. The reactant that has the smallest ratio is the limiting reactant.
For Mg(NO3)2:
Moles of Mg(NO3)2 / Coefficient of Mg(NO3)2 = 0.674 mol / 3 = 0.225 mol
For Li3P:
Moles of Li3P / Coefficient of Li3P = 0.864 mol / 2 = 0.432 mol
As we can see, Mg(NO3)2 has a smaller ratio, so it is the limiting reactant.
Step 3: Calculate the amount of products formed.
From the balanced equation, we see that the molar ratio of Mg(NO3)2 to Mg3P2 is 3:2. Since Mg(NO3)2 is the limiting reactant, we use its moles to determine the moles of Mg3P2 produced.
Moles of Mg3P2 = (0.674 mol of Mg(NO3)2) x (2 mol of Mg3P2 / 3 mol of Mg(NO3)2)
= 0.449 mol
Moles of LiNO3 = (0.449 mol of Mg(NO3)2) x (6 mol of LiNO3 / 3 mol of Mg(NO3)2)
= 0.898 mol
Step 4: Convert the moles of products to grams.
Mass of Mg3P2 = (moles of Mg3P2) x (molar mass of Mg3P2)
= 0.449 mol x 134.83 g/mol
= 60.54 g
Mass of LiNO3 = (moles of LiNO3) x (molar mass of LiNO3)
= 0.898 mol x 68.95 g/mol
= 61.74 g
Step 5: Calculate the amount of reactant remaining.
To find the amount of reactant remaining, we need to subtract the moles consumed from the moles initially present.
Moles of Mg(NO3)2 Remaining = Initial moles of Mg(NO3)2 - Moles consumed
= 0.674 mol - 0.449 mol
= 0.225 mol
Moles of Li3P Remaining = Initial moles of Li3P - Moles consumed
= 0.864 mol - 0.432 mol
= 0.432 mol
Step 6: Convert the moles of reactants remaining to grams.
Mass of Mg(NO3)2 Remaining = (moles of Mg(NO3)2 remaining) x (molar mass of Mg(NO3)2)
= 0.225 mol x 148.31 g/mol
= 33.37 g
Mass of Li3P Remaining = (moles of Li3P remaining) x (molar mass of Li3P)
= 0.432 mol x 115.79 g/mol
= 49.98 g
So, at the end of the reaction, you would obtain 60.54 grams of magnesium phosphide (Mg3P2), 61.74 grams of lithium nitrite (LiNO3), and have 33.37 grams of magnesium nitrite (Mg(NO3)2) and 49.98 grams of lithium phosphide (Li3P) remaining.