Sb2S3 + 3Fe --> 2Sb + 3FeS When 0.022mol Sb2S3 react with an excess of Fe, 0.08mol Sb is produced. What is the percent yield of
this reaction? (hint: what's the theoretical yield?)

The actual value is greater than the theoretical value

Given

Reaction

Sb2S3 + 3Fe --> 2Sb + 3FeS

0.022 mol Sb2S3

0.08mol Sb

Required

The percent yield

Solution

Percent yield is the comparison of the amount of product obtained from a reaction(actual) with the amount you calculated (theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

Theoretical yield of Sb : 2/1 x 0.022 mol = 0.044 moles

In the question, there seems to be an error, because the actual yield value(0.08=?) should be smaller than the theoretical yield(0.044)

your answer would be a. 12.5 kg

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.