Viagnesiumi anu If I have 100g of Magnesium, how much Magnesium Nitride will I theoretically create?
O 24.3g Mg3 N2
O 138.4g Mg3 N2
415.2g Mg3 N2
O 200g Mg3 N2

Theoretical yield is 138.4 g

Explanation:

In the first step we determine the reaction:

3Mg + N₂ → Mg₃N₂

Mass of reactant is 100 g. We assume the nitrogen is in excess, so we work with Mg. We convert mass to moles:

100 g . 1mol/ 24.3g = 4.11 moles of Mg.

Ratio is 3:1. 3 moles of Mg can produce 1 mol of nitride

Our 4.11 moles, may produce (4.11 . 1)/3 = 1.37 moles of Mg₃N₂

We convert mass to moles, to find the theoretical yield:

1.37 mol . 100.9 g/mol = 138.2 g

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

it is the most commonly accepted one