Question

3H2(g)+N2(g)?2NH3(g)

How many grams of NH3 can be produced from 3.23mol of N2 and excess H2.

How many grams of H2 are needed to produce 13.85g of NH3?

How many molecules (not moles) of NH3 are produced from 6.68

Answer 1

3H2(g)+N2(g)?2NH3(g)

1) 1 mole of N2 produces 2 moles of NH3

=> 3.23mol of N2 will produce = 3.23*2 =6.46 moles of NH3

mass of NH3 produced = no. of mole * molecular mass of NH3 = 6.46*17 = 109.82 g

2) moles of NH3 = given mass / molecular mass = 13.85 / 17 = 0.815 moles

2 moles of NH3 is produced from 3 moles of H2

=> 1 mole of NH3 is produced from 3/2 = 1.5 moles of H2

=> 0.815 mole of NH3 is produced from 1.5*0.815 = 1.22 moles of H2

mass of H2 = no. of mole * molecular mass of H2 = 1.22*2 = 2.44 g

3) no. of moles of H2 = given mass / molecular mass = 6.68*10^?4 / 2 = 3.34 *10^?4 moles

3 moles of H2 produces 2 moles of NH3

=>1 mole of H2 produces 2/3 moles of NH3

=> 3.34 *10^?4 moles of H2 produces (2/3)*3.34 *10^?4 =2.227*10^-4 moles of NH3

1 mole of any substance contain avagadro no. of molecules = 6.022*10²³ molecules

=> 2.227*10^-4 moles of NH3 will have 2.227*10^-4 * 6.022*10²³ = 1.34*10²⁰ molecules