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)

"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,

→

      (produced)

hence,

→ Mass of CO₂ = Moles × Molar mass of CO₂

                         =

                         =

Thus the above approach is right.

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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

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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~ trueboss72