Fool's gold is not gold at all but only lead IV sulfide. If you have 751.0 grams of lead how many grams of fool's gold can you make?
Ans: Pb+2S->PbS2
Given * (1/amu of given) * (mol of find/mol of given) * (amu of find/1mol)
Convert 751 g Pb to mols. mols = grams/atomic mass Pb.
Convert mols Pb to mols PbS2 using the coefficients in the balanced equation.
Now convert mols PbS2 to grams. g = mols x molar mass.
To calculate how many grams of fool's gold (lead IV sulfide) can be made from 751.0 grams of lead, we will use the balanced chemical equation for the reaction:
Pb + 2S -> PbS2
From the equation, we can see that the molar ratio between lead (Pb) and lead IV sulfide (PbS2) is 1:1.
1. Convert the given mass of lead (Pb) to moles:
Molar mass of Pb = 207.2 g/mol
Moles of Pb = (given mass of Pb) / (molar mass of Pb)
= 751.0 g / 207.2 g/mol
≈ 3.626 mol
2. Use the molar ratio from the balanced equation to calculate the moles of lead IV sulfide (PbS2):
Moles of PbS2 = Moles of Pb
= 3.626 mol
3. Convert the moles of lead IV sulfide (PbS2) to grams:
Molar mass of PbS2 = molar mass of Pb + (2 * molar mass of S)
= 207.2 g/mol + (2 * 32.1 g/mol)
= 271.4 g/mol
Grams of PbS2 = Moles of PbS2 * (molar mass of PbS2)
= 3.626 mol * (271.4 g/mol)
≈ 983.4 g
Therefore, from 751.0 grams of lead, you can make approximately 983.4 grams of fool's gold (lead IV sulfide).