Question

Mg3N2 + H2O + Mg(OH)2 + NH,
2NH3 + 02 2N0 + 3H20
What amount of M93N2 must be used to produce sufficient ammonia so that it may give 400 g of NO?

Answer 1

Mg3N2 + H2O + Mg(OH)2 + NH,

2NH3 + 02+ 2N0 + 3H20

now we know that,

moles of reactant/stotiometry=moles of pdts/stotiometry

mass of NO=400g

molar mass of NO=30g

moles NO=13.3

Moles(Mg3N2)/1=Moles (NO)/2

Moles(Mg3N2)=6.66

Mass of (Mg3N2)=100

mass= moles×molar mass

mass=666g

I hope this helps ( ╹▽╹ )