Find the moles of each product formed when 3.40 mol c3h7oh reacts with oxygen

the balanced equation for the combustion reaction is as follows

2C₃H₇OH + 9O₂ ---> 6CO₂ + 8H₂O

we are assuming that O₂ is in excess and C₃H₇OH is the limiting reactant

therefore the amount of product formed depends on the amount of C₃H₇OH present.

we have to find the number of CO₂ and H₂O moles formed

molar ratio of C₃H₇OH to CO₂ is 2:6 simplified to 1:3

when 1 mol of C₃H₇OH reacts - 3 mol of CO₂ is formed

therefore when 3.40 mol of C₃H₇OH reacts - 3 x 3.40 = 10.2 mol of CO₂ is formed

molar ratio of C₃H₇OH to H₂O is 2:8 simplified to 1:4

when 1 mol of C₃H₇OH reacts - 4 mol of H₂O is formed

therefore when 3.40 mol of C₃H₇OH reacts - 4 x 3.40 = 13.6 mol of H₂O is formed

10.2 mol of CO₂ and 13.6 mol of H₂O is formed

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.