If 16.00 g of O₂ reacts with 80.00 g NO, how many grams of NO₂ are produced? (enter only the value, round to whole number)

46 g

Explanation:

The balanced equation of the reaction between O and NO is

2 NO  +  O₂  ⇔  2 NO₂

Now, you need to find the limiting reagent.  Find the moles of each reactant and divide the moles by the coefficient in the equation.

NO:  (80 g)/(30.006 g/mol) = 2.666 mol

        (2.666 mol)/2 = 1.333

O₂:  (16 g)/(31.998 g/mol) = 0.500 mol

     (0.500 mol)/1 = 0.500 mol

Since O₂ is smaller, this is the limiting reagent.

The amount of NO₂ produced will depend on the limiting reagent.  You need to look at the equation to determine the ratio.  For every mole of O₂ reacted, 2 moles of NO₂ are produced.

To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂.  Then, convert moles of NO₂ to find grams.

0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂

1.000 mol × 46.005 g/mol = 46.005 g

You will produce 46 g of NO₂.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.