Question

The mass of mg3 n2 produced if 48gm of mg metal is reacted with 34gm nh3 gas is

Answer 1

Hey----

Mg = 24

Mg3N2 = 24x3 + 14x2 = 100g/mol

NH3 = 14 + 1x3 = 17g/mol

Moles of Mg = 48/24 = 2

Moles of NH3 = 34/17 = 2

From the equation, 

3Mg+2NH3 = Mg3N2 + 3H2

2 moles of NH3 react with 3 moles of Mg

Implies that NH3 is the limiting reagent and that 1 mole of Mg3N2 will be produced

Mass of Mg3N2 produced = Molar mass x number of moles

= 100 x 1

= 100g