Question

According to the reaction: $Pb\left(s\right)+S\left(s\right)\to PbS\left(s\right)$, when 20.7 grams of lead are reacted with 6.4 grams of sulfur :

• A. there will be an excess of 20.7 grams 04 lead
• B. the sulfur will be in excess by 3.2 grams
• C. the lead and sulfur will react completely without any excess reactants
• D. the sulfur will be the limiting factor in the reaction
• E. there will be an excess of 10.35 grams of lead

Answer 1

The correct option is B the sulfur will be in excess by 3.2 grams

The atomic masses of Pb and S are 207 g/mol and 32 g/mol.
The number of moles is the ratio of mass to molar mass.
The number of moles of Pb $=\frac{20.7g}{207g/mol}=0.1$ mole.
The number of moles of S $=\frac{6.4g}{32g/mol}=0.2$ mol.
Thus S is excess reactant and Pb is limiting reactant.
Out of 0.2 moles of S, 0.1 moles will react with 0.1 moles of Pb and 0.1 moles of S will be the excess reactant. The mass of excess S $=0.1mol\times 32g/mol=3.2g$.
Thus, the sulfur will be in excess by 3.2 grams.