mols Sb2S3 = 3.87E23/6.02E23 = ?
Then mols Sb2S3 x (3 mols FeS/1 mol Sb2S3) = ?
Finally, grams = mols x molar mass
Given the equation below, if 3.87×1023 particles of Sb2S3(s) are reacted with excess Fe(s), what mass of FeS(s) is produced?
Sb2S3(s) + 3Fe(s)→2Sb(s) +3FeS(s)
1 answer