How many grams of H2S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH3COO)2?

2.56g of H₂S

Explanation:

The reaction of H₂S with Pb(CH₃COO)₂ is:

H₂S + Pb(CH₃COO)₂ → 2 CH₃COOH + PbS

Where 1 mole of H₂S produce 1 mole of PbS

18.00g of PbS (Molar mass: 239.3g/mol):

18.00g of PbS ₓ (1mol / 239.3g) = 0.0752 moles of PbS

As 1 mole of PbS is produced from 1 mole of H₂S, moles needed of H₂S are 0.0752

As molar mass of H₂S is 34.1g/mol, mass in 0.0752moles is:

0.0752 mol H₂S ₓ (34.1g / mol) = 2.56g of H₂S

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.