find mass of magnesium nitride produced in reaction of 5.0of magnesium with 10.0gof nitrogen. the reaction was carried out in a closed container
See your other limiting reagent problems.
your not getting the picture I need help to study for an exam because I do not get this.
To find the mass of magnesium nitride produced in the reaction, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and restricts the amount of product that can be formed.
To do this, we need to calculate the moles of each reactant using their molar masses. The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of nitrogen (N₂) is 28.02 g/mol.
1. Calculate the moles of magnesium (Mg):
moles of Mg = mass of Mg / molar mass of Mg
= 5.0 g / 24.31 g/mol
2. Calculate the moles of nitrogen (N₂):
moles of N₂ = mass of N₂ / molar mass of N₂
= 10.0 g / 28.02 g/mol
3. Determine the limiting reactant:
The limiting reactant is the one with fewer moles. Compare the moles of Mg and N₂ calculated above.
4. To proceed, let's assume that magnesium (Mg) is the limiting reactant.
5. Write the balanced chemical equation for the reaction:
3 Mg + N₂ → Mg₃N₂
6. From the balanced equation, we can see that three moles of magnesium react with one mole of nitrogen to produce one mole of magnesium nitride.
7. Calculate the moles of magnesium nitride (Mg₃N₂) that can be formed from the limiting reactant (magnesium, assuming it is all consumed):
moles of Mg₃N₂ = moles of Mg / 3
8. Finally, calculate the mass of magnesium nitride (Mg₃N₂) produced:
mass of Mg₃N₂ = moles of Mg₃N₂ x molar mass of Mg₃N₂
Please note that step 3 involves comparing the moles of magnesium and nitrogen to determine the limiting reactant. However, without knowing the balanced chemical equation, it is not possible to provide an exact calculation. The steps provided above are a general approach to solve such stoichiometry problems.