Question

При нагреваний железа (28г)с серой(32г)образуется сульфид железа массой (г)

Answer 1

$Fe + S = FeS$

Mr(Fe) = 56 г/моль
m(Fe) = 56 * 1 = 56 г
v(Fe) = 28/56 = 0,5 моль
Mr(S) = 32 г/моль
m(S) = 32 * 1 = 32 г
v(S) = 32/32 = 1 моль
сера в избытке, значит расчет ведем по железу
$m(FeS)= \frac{28 * 88}{56}=44$ г

Ответ: m(FeS) = 44 г

Answer 2

Calculating the Mass of Iron(II) Sulfide Formed

To calculate the mass of iron(II) sulfide formed when iron (28g) reacts with sulfur (32g), we can use the concept of stoichiometry and the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between iron and sulfur to form iron(II) sulfide is: Fe + S → FeS

According to the law of conservation of mass, the total mass of the reactants should be equal to the total mass of the products.

Using Stoichiometry to Calculate the Mass of Iron(II) Sulfide

To calculate the mass of iron(II) sulfide formed, we can use the molar masses of iron and sulfur to determine the limiting reactant and then use stoichiometry to find the mass of iron(II) sulfide formed.

The molar mass of iron (Fe) is approximately 55.85 g/mol, and the molar mass of sulfur (S) is approximately 32.07 g/mol.

First, we need to determine the limiting reactant by comparing the moles of each reactant to the stoichiometric coefficients in the balanced chemical equation.

The moles of iron can be calculated using the given mass of iron (28g) and its molar mass: Moles of Fe = Mass of Fe / Molar mass of Fe = 28g / 55.85 g/mol

The moles of sulfur can be calculated using the given mass of sulfur (32g) and its molar mass: Moles of S = Mass of S / Molar mass of S = 32g / 32.07 g/mol

After determining the limiting reactant, we can use stoichiometry to calculate the mass of iron(II) sulfide formed.

Calculation

Let's calculate the moles of iron and sulfur and determine the limiting reactant.