Question

Для реакции взято 14г. железа и 10г. серы. Считая, что реакция идет до конца,вычислите массу сульфида железа( II ).

Answer 1

  14г      10г
  Fe  +   S   =   FeS
1моль 1моль

задача на избыток и недостаток. т.к. даны массы 2-х веществ вступающих в реакцию
найдем количества железа и серы. кол-во = масса / молярную ассу
v (Fe) = 14/56 = 0.25 моль
v (S) = 10/32 = 0.313 моль
по уравнению реакции железа и сера находятся в одномолярных отношениях, значит 0,25 моль железа должно прореагировать с 0,25 моль серы (а ее 0,313)
значит сера в избытке, расчет ведем по недостатку - железу.
кол-во сульфида = кол-ву железа
v(FeS)  =  v (Fe) = 0.25моль
m (FeS) = v * M = 0.25 * 88 =22г

Answer 2

Calculating the Mass of Iron(II) Sulfide (FeS)

To calculate the mass of iron(II) sulfide (FeS) produced in the reaction, we can use the stoichiometry of the reaction and the given masses of iron and sulfur.

Given: - Mass of iron (Fe) = 14g - Mass of sulfur (S) = 10g

The balanced chemical equation for the formation of iron(II) sulfide is: 2Fe + S → FeS

From the balanced equation, we can see that 2 moles of iron react with 1 mole of sulfur to produce 1 mole of iron(II) sulfide.

Using the molar masses: - Molar mass of Fe = 55.85 g/mol - Molar mass of S = 32.07 g/mol

We can calculate the moles of iron and sulfur: - Moles of Fe = Mass of Fe / Molar mass of Fe - Moles of S = Mass of S / Molar mass of S

Then, we can determine the limiting reactant and use it to calculate the mass of iron(II) sulfide produced.

Let's proceed with the calculations.

Step 1: Calculate the moles of iron and sulfur - Moles of Fe = 14g / 55.85 g/mol - Moles of S = 10g / 32.07 g/mol

Step 2: Determine the limiting reactant - The limiting reactant is the reactant that is completely consumed in the reaction. It can be determined by comparing the moles of each reactant to the stoichiometry of the balanced equation.

Step 3: Calculate the moles of FeS produced - Since the reaction goes to completion, the moles of FeS produced will be determined by the limiting reactant.

Step 4: Convert moles of FeS to grams - Moles of FeS × Molar mass of FeS = Mass of FeS

Let's perform the calculations.

Results:

- Moles of Fe = 14g / 55.85 g/mol = 0.2501 moles - Moles of S = 10g / 32.07 g/mol = 0.3118 moles

Since iron is the limiting reactant (as it has fewer moles than sulfur), we will use the moles of iron to calculate the moles of FeS produced.

- Moles of FeS produced = 0.2501 moles - Molar mass of FeS = 55.85 g/mol + 32.07 g/mol = 87.92 g/mol - Mass of FeS = 0.2501 moles × 87.92 g/mol = 21.99g

The mass of iron(II) sulfide (FeS) produced in the reaction is approximately 21.99 grams.