Question

количество вещества (моль) углекислого газа получится при реакции 0,3 моль серной кислоты и 0,1 моль углерола​

Answer 1

Дано:

n₁(H₂SO₄)=0,3моль

n₁(C)=0,1моль

----------------------------

n₁(CO₂)-?

Решение:

1. C+2H₂SO₄(конц. t°C)=CO₂+2SO₂+2H₂O

по уравнению реакции:

n(C)n(H₂SO₄)=1моль  n(H₂SO₄)=2моль n₁(CO₂)=1моль

по условию задачи:

n(C)n(H₂SO₄)=0,1моль  n(H₂SO₄)=0,3моль n₁(CO₂)=0,15моль

2. Ответ: при реакции 0,3 моль серной кислоты и 0,1 моль углерода​ образуется 0,15моль углекислого газа.

Answer 2

To determine the amount of carbon dioxide (CO2) produced during the reaction between 0.3 moles of sulfuric acid (H2SO4) and 0.1 moles of carbon monoxide (CO), we need to write and balance the chemical equation for the reaction.

The balanced chemical equation for the reaction between sulfuric acid and carbon monoxide is as follows:

H2SO4 + 6CO → 6CO2 + H2S

From the balanced equation, we can see that 1 mole of sulfuric acid (H2SO4) reacts with 6 moles of carbon monoxide (CO) to produce 6 moles of carbon dioxide (CO2).

Now, let's calculate the amount of carbon dioxide (CO2) produced:

Given:

• Moles of sulfuric acid (H2SO4) = 0.3 moles
• Moles of carbon monoxide (CO) = 0.1 moles

Since 1 mole of H2SO4 reacts with 6 moles of CO, we need to determine the limiting reactant to find out how much CO2 will be produced. The reactant that produces fewer moles of CO2 will be the limiting reactant.

Let's calculate the moles of CO2 produced by both reactants:

For sulfuric acid (H2SO4): Moles of CO2 = 0.3 moles H2SO4 * (6 moles CO2 / 1 mole H2SO4) = 1.8 moles CO2

For carbon monoxide (CO): Moles of CO2 = 0.1 moles CO * (6 moles CO2 / 6 moles CO) = 0.1 moles CO2

Now, we compare the moles of CO2 produced by each reactant. We see that carbon monoxide (CO) produces fewer moles of CO2 (0.1 moles), making it the limiting reactant.

Therefore, the amount of carbon dioxide (CO2) produced during the reaction is 0.1 moles.