Question

1. Calculate the number of molecules of oxygen in a 3.40 g of the gas

2. Calculate the mass of 3.543 x 10^23 atoms of magnesium

A. Balance the equation below then calculate the masses of carbon dioxide and water formed by burning 5.00 of methane in 5.00 g of oxygen: CH4(g) + O2(g) --> CO2(g) + H2O (l)

b) This is a very important reaction. Why?

Answer 1

1. no of molecules = W*6.023*10^23/G.M.Wt

                               = 3.4*6.023*10^23/32

                               = 6.4*10^22 molecules

2. no of atoms   = W* 6.023*10^23/G.A.Wt

     3.543*10^23    = W*6.023*10^23/24

          W                = 3.543*10^23 *24/6.023*10^23   = 14.12g

CH4(g) + 2O2(g) --> CO2(g) + 2H2O (l)

no of moles of CH4 = W/G.M.Wt   = 5/16 = 0.3125 moles

no of moles of O2   = W/G.M.Wt    = 5/32 = 0.15625 moles

from balanced equation

1 mole of CH4 react with 2 mole of O2

0.3125 moles of CH4 react with = 2*0.3125/1   = 0.625 moles of O2

O2 is limiting reagent

2 moles of O2 react with CH4 to form 1 mole of CO2

0.15625 moles of O2 react with CH4 to form = 1*0.15625/2    = 0.078125moles of Co2

mass of CO2 = no of moles * gram molar mass

                      = 0.078125*44   = 3.4375g