the element nitrogen and magnesium react to give magnesium nitride according to the balanced equation
N2 + 3Mg = Mg3N2
suppose you have 0.3616 mol of n2
1-how many moles of mg are needed for complete reaction?
2-what mass of Mg3N2, in grams can be produced? ( the molar mass of Mg3N2 is 100.9284 g/mol)
- please show step to step and the answere please becuase i need to go check it and see how you did it
thanks
Here is a worked example of a stoichiometry problem. Just follow the steps. http://www.jiskha.com/science/chemistry/stoichiometry.html
Sure! Let's go through the steps to find the answers to your questions.
1. To find the number of moles of Mg needed for complete reaction, we can use the mole ratio between N2 and Mg in the balanced equation. According to the equation, 1 mole of N2 reacts with 3 moles of Mg.
Given that you have 0.3616 mol of N2, we can use the mole ratio to find the number of moles of Mg needed.
Moles of Mg = (Moles of N2) x (Mole ratio)
= 0.3616 mol N2 x (3 mol Mg / 1 mol N2)
= 1.0848 mol Mg
Therefore, you would need 1.0848 moles of Mg for complete reaction.
2. To find the mass of Mg3N2 that can be produced, we need to use the stoichiometry of the balanced equation. From the balanced equation, we can see that 1 mole of N2 reacts to form 1 mole of Mg3N2.
Given that the molar mass of Mg3N2 is 100.9284 g/mol, we can calculate the mass of Mg3N2 produced.
Mass of Mg3N2 = (Moles of Mg3N2) x (Molar mass of Mg3N2)
= 1.0848 mol Mg3N2 x (100.9284 g / 1 mol Mg3N2)
= 109.401 g
Therefore, you can produce 109.401 grams of Mg3N2.
I hope this helps! Let me know if you have any more questions.