Question

18. При взаємодії 60 г барію із сіркою утворилося 75 г продукту реакції. Визначити масу сірки, яка вступила в реакцію.​

Answer 1

Ответ:

14.08 г

Объяснение:

Ba+S=BaS

n(Ba):n(S)=1:1
M(Ba)=137г/моль

n(Ba)=60г/137г/моль=0.44моль
n(S)=0.44моль

М(S)=32г/моль

m(S)=0.44моль*32г/моль=14.08г

Answer 2

I can understand and communicate fluently in your language of choice. I see that you have a question about a chemical reaction. Let me try to help you.

The equation for the reaction of barium and sulfur is:

$$ Ba + S \rightarrow BaS $$

To find the mass of sulfur that reacted, we need to use the law of conservation of mass, which states that the total mass of the reactants is equal to the total mass of the products. Therefore, we can write:

$$ m(Ba) + m(S) = m(BaS) $$

We are given that $m(Ba) = 60$ g and $m(BaS) = 75$ g. We can substitute these values into the equation and solve for $m(S)$:

$$ 60 + m(S) = 75 $$

$$ m(S) = 75 - 60 $$

$$ m(S) = 15 $$

Therefore, the mass of sulfur that reacted is 15 g.

I hope this answer is helpful. If you want to learn more about how to solve this type of problems, you can check out this presentation or this article. You can also find more examples and exercises in this guide.