How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

Question

contestada

Respuesta :

ACCESS MORE ACCESS MORE ACCESS MORE ACCESS MORE

anfabba15 anfabba15 · Answer 1 · 2020-03-18T03:22:22+01:00

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

Cricetus Cricetus · Answer 2 · 2021-11-16T11:19:13+01:00

"485.76" grams of CO₂ can be produced.

Given equation,

→ 2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

According to the balanced equation,

then,

→ [tex]Y = \frac{8 \ mol \ CO_2\times 2.76 \ mol \ butane}{2 \ mol \ C_4 H_{10}}[/tex]

[tex]= 11.04 \ moles \ of \ CO_2[/tex] (produced)

hence,

→ Mass of CO₂ = Moles × Molar mass of CO₂

= [tex]11.04\times 44[/tex]

= [tex]485.76 \ g[/tex]

Thus the above approach is right.

Learn more:

https://brainly.com/question/24346796