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mg3n2 + 3k2o --> 3mgo + 2k3n

if 6 moles of potassium oxide reacts with an excess amount of magnesium nitride, how many grams of potassium nitride will be made?

There will be 525.2 grams of K3N produced

Explanation:

Step 1: Data given

Number of moles of potassium oxide ( K2O) = 6 moles

Magnesium nitride (Mg3N) = in excess

Molar mass of K3N = 131.3 g/mol

Step 2: The balanced equation

Mg3N2 + 3K2O → 3MgO + 2K3N

Step 3: Calculate moles of K3N

The limiting reactant is K2O.

For 1 mol Mg3N2 consumed, we need 3 moles of K2O to produce 3 moles of MgO and 2 moles of K3N

For 6 moles K2O we'll have 2/3 * 6 = 4 moles of K3N

Step 4: Calculate mass of K3N

Mass of K3N = moles K3N * molar mass K3N

Mass of K3N = 4 moles * 131.3 g/mol

Mass of K3N = 525.2 grams

There will be 525.2 grams of K3N produced

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

the correct answer to this question is 2