Question

I need help with part B

21. a) Assume the following reaction was completed in the laboratory: 2 Sb(s)+3 FeS(s). Sb2S3(s)+ 3 Fe (s) 3.87 x 1023 particles of Sb2 S3 reacts with 25.0 g of Fe, how many grams of FeS will be produced? (8 points)

Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 1: write the balanced chemical equation.

Sb₂S₃(s) + 3Fe(s)→2Sb(s) +3FeS(s)

Step 2: calculate the moles of Sb₂S_{3 ,} Fe

Moles of Fe = mass given / molar mass = ( 25 g / 55.845 g/mol ) = 0.4476676515 mol

Moles of Sb₂S₃ = given particles / avogadro's number =   ( 3.87 × 10²³ particles  / 6.022 × 10²³ particles / mol ) = 0.643 mol

Step 3: Determine the limiting reagent

Sb₂S₃(s) + 3Fe(s)→2Sb(s) +3FeS(s)

According to the reaction we need
3 mol of Fe requires for 1 mol of Sb₂S₃
so, for 0.4476676515 mol of Fewe will need =(1 mol of Sb₂S₃ / 3 mol of Fe )× 0.4476676515 mol of Fe = 0.149 mol of Sb₂S₃
Since we need only 0.149 mol of Sb₂S₃ hence Sb₂S₃ is in excess ( since given =0.643 mol ) and  Fe is limiting reagent.

so the amount of product will formed according to the limiting reagent ( i.e Feamount )

Step 4: Calculate the moles of silver FeS produced

Sb₂S₃(s) + 3 Fe(s)→2 Sb(s) +3 FeS(s)

According to the reaction:
3 mol of Fe produce 3 mol of FeS
so, 0.4476676515 mol of Fe will produce=( 3 mol of FeS / 3 mol of Fe )× 0.4476676515 mol of Fe = 0.4476676515 mol of FeS

Step 5: Calculation of mass of FeS produced

we get moles of FeSthat can be produced = 0.4476676515 mol

Mass of FeSproduced =( moles×molar mass ) = ( 0.4476676515 mol × 87.91 g/mol ) = 39.3545 g

Answer 1

no of moles of Fe     = W/G.A.Wt

                                = 25/55.845    = 0.448moles

6.023*10^23 particles   = 1 mole of Sb2S3

3.87*10^23 particles   = 1*3.87*10^23/(6.023*10^23)   = 0.6425 moles of Sb2S3

Sb₂S₃(s) + 3Fe(s)→2Sb(s) +3FeS(s)

1 mole of Sb2S3 react with 3 moles of Fe

0.6425 moles of Sb2S3 react with = 3*0.6425/1   = 1.9275 moles of Fe is required

Fe is limiting reactant

3 moles of Fe react with excess of Sb2S3 to gives 3 moles of FeS

0.447 moles of Fe react with excess of Sb2S3 to gives = 3*0.448/3   = 0.448 moles of FeS

mass of FeS   = no of moles * gram molar mass

                         = 0.448*87.91   = 39.4g of FeS

Theoretical yield of FeS = 39.4g

actual yield of FeS   = 35g

percent yield = actual yield*100/theoretical yield

                       = 35*100/39.4   = 88.83% >>>>answer