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?
To determine the amount of each product and the leftover reactants in this chemical reaction, we need to first balance the equation for the reaction:
3 Mg(NO2)2 + 2 Li3P -> Mg3P2 + 6 LiNO2
The balanced equation tells us that 3 moles of magnesium nitrite react with 2 moles of lithium phosphide to produce 1 mole of magnesium phosphide and 6 moles of lithium nitrite.
To find the number of moles of each reactant, we need to convert the mass of the reactants into moles using their respective molar masses. The molar masses are as follows:
- Magnesium nitrite (Mg(NO2)2): 148.3 g/mol (24.3 g/mol for Mg + 46.0 g/mol for NO2)
- Lithium phosphide (Li3P): 103.8 g/mol (3 * 6.9 g/mol for Li + 30.0 g/mol for P)
Now let's calculate the number of moles for both reactants:
For magnesium nitrite:
Number of moles = Mass / Molar mass
Number of moles = 100 g / 148.3 g/mol
Number of moles = 0.674 mol
For lithium phosphide:
Number of moles = Mass / Molar mass
Number of moles = 100 g / 103.8 g/mol
Number of moles = 0.962 mol
Next, we need to determine the limiting reactant, which is the reactant that will be consumed completely or limit the amount of product formed. The limiting reactant is the one that produces the fewer moles of product based on the balanced equation.
From the balanced equation, the ratio between magnesium nitrite and magnesium phosphide is 3:1. So, based on the moles we calculated:
3 moles of Mg(NO2)2 produce 1 mole of Mg3P2, and we have 0.674 moles of Mg(NO2)2.
0.674 moles of Mg(NO2)2 * (1 mole of Mg3P2 / 3 moles of Mg(NO2)2) = 0.225 moles of Mg3P2
From the balanced equation, the ratio between lithium phosphide and magnesium phosphide is 2:1. So, based on the moles we calculated:
2 moles of Li3P produce 1 mole of Mg3P2, and we have 0.962 moles of Li3P.
0.962 moles of Li3P * (1 mole of Mg3P2 / 2 moles of Li3P) = 0.481 moles of Mg3P2
We can see that we have fewer moles of magnesium phosphide (Mg3P2) from both reactants, so it is the limiting reactant.
To calculate the moles of the excess reactants remaining after the reaction, we subtract the moles of limiting reactant from the initial moles of each reactant:
For magnesium nitrite:
Excess moles = Initial moles - Moles used
Excess moles = 0.674 mol - 0.225 mol
Excess moles = 0.449 mol
For lithium phosphide:
Excess moles = Initial moles - Moles used
Excess moles = 0.962 mol - 0.481 mol
Excess moles = 0.481 mol
Finally, we convert the excess moles of each reactant back into grams using their respective molar masses:
For magnesium nitrite:
Mass = Excess moles * Molar mass
Mass = 0.449 mol * 148.3 g/mol
Mass = 66.63 g
For lithium phosphide:
Mass = Excess moles * Molar mass
Mass = 0.481 mol * 103.8 g/mol
Mass = 49.97 g
Therefore, at the end of the reaction, you would produce:
- 0.225 grams of magnesium phosphide (Mg3P2)
- 66.63 grams of excess magnesium nitrite (Mg(NO2)2)
- 49.97 grams of excess lithium phosphide (Li3P)