If 16.00 g of O₂ reacts with 80.00 g NO, what is the excess reactant? *

A. O₂
B. NO

  B. NO

Explanation:

In each mole of NO, the weight of the oxygen is 16.00 g; the weight of the nitrogen is 14.01 g. Then in 80 g of NO, we have 42.65 g of oxygen and 37.35 g of nitrogen.

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In each mole of NO₂, the weight of the oxygen is 2(16.00) = 32 g. The weight of the nitrogen is 1(14.01) = 14.01 g.

From the available oxygen, we can produce ...

 (42.65 g +16 g)/(32.00 g/mol) = 1.83 mol of NO₂

From the available nitrogen, we can produce ...

 (37.35 g)/(14.01 g/mol) = 2.67 mol of NO₂

Clearly, the reaction is limited by the amount of available oxygen.

NO is the excess reactant.

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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.

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year