Nitric monoxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO₂), a dark brown gas. If 5.895 mol of NO is mixed with 2.503 mol of O₂, determine the limiting reagent.

calculate the number of grams of NO₂ produced.

and determine how many grams of excess reagent remain unreacted.

Limiting reactant: O2

grams NO2 produced = 230.276 g NO2

grams of NO unused = 26.67 gNO

Explanation:

2NO + O2 --> 2NO2

Step 1: Determine the molar ratio NO:O2

molar ratio NO:O2 = 5.895: 2.503 = 2.35

stoichiometric molar ratio NO:O2 = 2:1

So, O2 is the limiting reactant.

Step2: Determine the grams of NO2:

?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2

Step 3: Determine the amount of excess reagent unreacted

moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted

moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted

mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted

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:

if you mean hydrogen and oxygen then water is the product