Question

У разі взаємодії сірки масою 16г із залізом утворився ферум (||) сульфід FeS масою 44г. Складіть рівняння цієї реакції та обчисліть масу заліза, що витратилась у результаті взаємодії. _____Помогите пожалуйста​

Answer 1

Ответ:

n(S) = m\M= 16 г\32 г\моль = 0,5 моль

M(FeS) = 56+32 = 88г\моль

n(FeS) = 44г \ 88 г\моль = 0,5 моль

х моль      0,5 моль

Fe +  S  =  FeS

1 моль         1 моль

х = 1 ∙ 0,5\1= 0,5 моль

M(Fe) = n·M = 0,5 моль · 56 г\моль = 28 г

Answer 2

Reaction Equation and Calculation of Iron Mass

The reaction between sulfur and iron can be represented by the following balanced chemical equation:

Fe + S → FeS

According to the given information, the mass of sulfur (S) is 16 g and the mass of iron sulfide (FeS) formed is 44 g.

To calculate the mass of iron (Fe) that reacted, we need to determine the difference in mass before and after the reaction.

The molar mass of sulfur (S) is 32 g/mol, and the molar mass of iron sulfide (FeS) is 88 g/mol (56 g/mol for iron + 32 g/mol for sulfur).

Let's calculate the number of moles of sulfur (S) and iron sulfide (FeS) formed:

Number of moles of sulfur (S): n(S) = mass / molar mass = 16 g / 32 g/mol = 0.5 mol [[1]]

Number of moles of iron sulfide (FeS): n(FeS) = mass / molar mass = 44 g / 88 g/mol = 0.5 mol [[1]]

Since the balanced equation shows a 1:1 ratio between sulfur and iron sulfide, we can conclude that 0.5 moles of iron reacted.

To calculate the mass of iron (Fe) that reacted, we can use the molar mass of iron (56 g/mol):

Mass of iron (Fe) that reacted: mass(Fe) = moles of iron (Fe) × molar mass of iron (Fe) = 0.5 mol × 56 g/mol = 28 g [[1]]

Therefore, the mass of iron that reacted in the reaction is 28 g.

I hope this helps! Let me know if you have any further questions.