Question

a) How many grams of NH₃ can be produced from 4.75 mol of N₂ and excess H_2?

b) How many grams of H₂ are needed to produce 11.85 g of NH_3?

c) How many molecules (not moles) of NH₃ are produced from 7.98x10^-4 g of H_2?

d) What mass of HCl in grams is neutralized by a dose of milk of magnesia containing 3.55 g Mg(OH)_2?

Answer 1

a) N₂ + 3H₂ ----------------- 2NH₃            NH₃= 14x2+1x6= 34 g

   4.75 mol------------------- x g

    1 mol   -------------------- 34g

X= 161.5 g of NH₃

b) N₂ + 3H₂ ----------------- 2NH₃     

             x g------------------- 11.85 g

             6 g ----------------- 34 g

X= 2.09 g of H₂

c) N₂ + 3H₂ ----------------- 2NH₃ remember 1 mole has 6.022x10²³ molecules

         7.98x10^-4g---------- x molecules

            6 g   ---------------- 1.2044x10²⁴ molecules

X= 1.60x10²⁰ molecules

d) we have to know how many moles are neutralized first:

moles=Mass/molecular weight

moles of Mg(OH)₂=3.55/58.3

moles = 0.0601

by every 1 Mg(OH)₂ we have 2 OH, the moles of OH are: 0.06 x 2=0.12 moles

this is the same number of moles of HCl neutralized:

Moles of Hcl= 0.12

Mass= MolesxMolecular weight

Mass= 0.12 x 36.5

Mass of HCl= 4,38 g