Balanced equation first.
3Mg + N2 -> Mg3N2
55.3 grams Mg (1 mole Mg/24.31 grams)(1 mole Mg3N2/3 mole Mg)(100.95 grams/1 mole Mg3N2)
= 76.5 grams Mg3N2 made
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Need formula of magnesium nitride, an ionic compound. Mg2+ and N3- together make Mg3N2 -------------- so 3 * 24.31 g = 72.93 grams 2 * 14.01 g = 28.02 grams ------------------------------------------add = 100 .95 grams/mole --------------------------------molar mass magnesium nitride
7,68 grams of calcium nitride is equal to 0,052 moles.
i dont know help :C
15 grams of nitrogen are equal to 1,071 moles.
Theoretically the mass is 62,3018 g.
Need formula of magnesium nitride, an ionic compound. Mg2+ and N3- together make Mg3N2 -------------- so 3 * 24.31 g = 72.93 grams 2 * 14.01 g = 28.02 grams ------------------------------------------add = 100 .95 grams/mole --------------------------------molar mass magnesium nitride
3Mg + N2 --> Mg3N2 molar mass of Mg3N2: 100.9494 g/molmolar mass of Mg: 24.305 1.19g Mg3N2 * 1mol/100.9494g * 3mol Mg/1mol Mg3N2 * 24.305g/1mol = 0.860g
Barium nitride is Ba3N2 (from the rules of ionic bonding). Its molar mass is 439 grams/mole (from the periodic table), so 18.8 grams of it is 0.0428 moles (by algebraic direct proportion). Multiply this by Avogadro's number (6.02x1023; the number of particles in 1 mole) and you get 2.58x1022 formula units of barium nitride, and there are 2 nitrogen atoms per formula unit, so double it to 5.16x1022 atoms of nitrogen.
13.5g Mg(NO3)2 x 1 mol Mg(NO3)2/148.3 = 0.0910 mol Mg(NO3)2
7,68 grams of calcium nitride is equal to 0,052 moles.
8 g of magnesium chloride is obtained.
2
1 mole of ammonium nitrate produces one mole of nitrogen. Actually the amount (in moles) of nitrogen will depend on how much NH4NO3 you are starting with, what other reactant you are combining it with and whether or not the NH4NO3 completely reacts. Since you will never be able to retrieve all of the nitrogen (either the NH4 or the NO3 will retain some nitrogen depending upon the reaction), you can reasonably expect to get 1 mole of N2 for each 14.01 grams of Ammonium nitrate that COMPLETELY reacts.
340 grams
i dont know help :C
15 grams of nitrogen are equal to 1,071 moles.
Theoretically the mass is 62,3018 g.