Calculate the number of grams of magnesium nitride necessary to produce 2.02 L of ammonia gas at STP.
I got 4.55g for the upper question which is correct but idk how to do the bottom one:
How many moles of magnesium hydroxide can also be formed?
Mg3N2(s) + 6 H2O(l)--> 3 Mg(OH)2(aq) + 2 NH3(g)
thanks
I find it amazing you can do the first part but not the second part. Both are the same kind of stoichiometry problem. Here is the first one, then to the second one. I want you to see the similarity.
2.02L at STP = 2.02L x (1 mol/22.4L) = 0.09018 moles NH3.
Using the coefficients in the balanced equation, convert moles NH3 to moles Mg3N2. 0.09018 moles NH3 x (1 mole Mg3N2/2 moles NH3) = 0.09018 x (1/2) = 0.04509 moles Mg3N2.
Now convert moles Mg3N2 to grams. g = moles x molar mass = 0.04509 x 100.93 = 4.55 g Mg3N2.
Since we had to determine moles NH3, we may as well start there. moles NH3 = 0.09018.
Using the coefficients in the balanced equation convert moles NH3 to moles Mg(OH)2. moles NH3 x (3 moles Mg(OH)2/2 moles NH3) = 0.09018 x (3/2) = 0.1357 moles Mg(OH)2.
Convert moles to grams by g = moles x molar mass. 0.1357 moles Mg(OH)2 x molar mass Mg(OH)2 = ??
To calculate the number of moles of magnesium hydroxide (Mg(OH)2) formed, we need to use stoichiometry. The balanced chemical equation shows that 3 moles of Mg(OH)2 are produced for every mole of Mg3N2 reacted.
Since you have already found the number of grams of magnesium nitride required (4.55 g), we can use the molar mass of Mg3N2 to convert it to moles. The molar mass of Mg3N2 is 100.95 g/mol.
Moles of Mg3N2 = mass / molar mass
Moles of Mg3N2 = 4.55 g / 100.95 g/mol
Moles of Mg3N2 = 0.045 moles
Now, we can use the stoichiometric ratio from the balanced chemical equation to determine the number of moles of Mg(OH)2 formed.
Moles of Mg(OH)2 = Moles of Mg3N2 × (3 moles of Mg(OH)2 / 1 mole of Mg3N2)
Moles of Mg(OH)2 = 0.045 moles × (3 / 1)
Moles of Mg(OH)2 = 0.135 moles
Therefore, 0.135 moles of magnesium hydroxide can be formed from the given amount of magnesium nitride.